Ideas: YCT 20. Parabola

YCT 20. Parabola

Type I Standard equation of Parabola

1. Let P be a parabola with vertex $(2, 3)$ and directrix $2x + y = 6$. Let an ellipse $E : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a > b$, of eccentricity $\frac{1}{2}$ pass through the focus of the parabola $P$. Then the square of the length of the latus rectum of $E$?

(a) $\frac{512}{25}$

(b) $\frac{385}{8}$

(d) $\frac{347}{8}$

JEE Mains 31/01/2024 Shift-II

Ans. (c) :

Foot of perpendicular from vertex $(2,3)$ to directrix $2x + y - 6 = 0$ is given by:

$$\frac{x - 2}{2} = \frac{y - 3}{1} = -\frac{2(2) + 3 - 6}{2^2 + 1^2} = -\frac{1}{5}$$ $$\Rightarrow x - 2 = -\frac{2}{5} \implies x = \frac{8}{5}$$ $$\Rightarrow y - 3 = -\frac{1}{5} \implies y = \frac{14}{5}$$

Vertex is the midpoint of the focus $(h, k)$ and the foot of directrix $\left(\frac{8}{5}, \frac{14}{5}\right)$:

$$\frac{h + \frac{8}{5}}{2} = 2 \implies h = 4 - \frac{8}{5} = \frac{12}{5}$$ $$\frac{k + \frac{14}{5}}{2} = 3 \implies k = 6 - \frac{14}{5} = \frac{16}{5}$$

Focus of parabola lies on ellipse $E$, so point $\left(\frac{12}{5}, \frac{16}{5}\right)$ satisfies $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$:

$$\frac{\left(\frac{12}{5}\right)^2}{a^2} + \frac{\left(\frac{16}{5}\right)^2}{b^2} = 1 \implies \frac{144}{25a^2} + \frac{256}{25b^2} = 1 \quad \text{---(1)}$$

Eccentricity $e = \frac{1}{2}$:

$$e^2 = 1 - \frac{b^2}{a^2} = \frac{1}{4} \implies \frac{b^2}{a^2} = \frac{3}{4} \implies a^2 = \frac{4}{3}b^2 \quad \text{---(2)}$$

Using eq. (2) in equation (1):

$$\frac{144}{25 \left(\frac{4}{3}b^2\right)} + \frac{256}{25b^2} = 1$$ $$\frac{108}{25b^2} + \frac{256}{25b^2} = 1 \implies \frac{364}{25b^2} = 1 \implies b^2 = \frac{328}{25}$$

Length of Latus Rectum of $E = \frac{2b^2}{a}$

Square of length of L.R. $= \frac{4b^4}{a^2} = \frac{4b^4}{\frac{4}{3}b^2} = 3b^2 = 2 \times b^2 = \frac{656}{25}$

2. Let a line perpendicular to the line $2x - y = 10$ touch the parabola $y^2 = 4(x - 9)$ at the point $P$. The distance of the point $P$ from the centre of the circle $x^2 + y^2 - 14x - 8y + 56 = 0$ ______.

JEE Mains 05/04/2024 Shift-II

Ans. (10) : Given equation of circle & parabola

$$x^2 + y^2 - 14x - 8y + 56 = 0$$ $$y^2 = 4(x - 9)$$

Given line $L_1 : 2x - y = 10$. Slope of line $L_1 = 2$.

A line perpendicular to $L_1$ has slope $m = -\frac{1}{2}$.

For parabola $y^2 = 4(x - 9)$, $a = 1$.

Equation of tangent with slope $m$ for $y^2 = 4(x - 9)$ is:

$$y = m(x - 9) + \frac{a}{m}$$ $$y = -\frac{1}{2}(x - 9) + \frac{1}{-\frac{1}{2}} = -\frac{1}{2}x + \frac{9}{2} - 2$$ $$y = -\frac{1}{2}x + \frac{5}{2} \implies x + 2y = 5$$

Point of contact $P(x_1, y_1)$ on standard parabola $y^2 = 4ax$ is $\left(\frac{a}{m^2}, \frac{2a}{m}\right) = (4, -4)$.

Shifting origin for $y^2 = 4(x - 9)$:

$$P = \left(9 + \frac{a}{m^2}, \frac{2a}{m}\right) = \left(9 + 4, -4\right) = (13, -4)$$

Center of circle $x^2 + y^2 - 14x - 8y + 56 = 0$ is $C(7, 4)$.

Distance $CP = \sqrt{(13 - 7)^2 + (-4 - 4)^2} = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = 10$.

3. Let $A, B$ and $C$ be three points on the parabola $y^2 = 6x$ and let the line segment $AB$ meet the line $L$ through $C$ parallel to the x-axis at the point $D$. Let $M$ and $N$ respectively be the feet of the perpendiculars from $A$ and $B$ on $L$. Then $\left(\frac{AM \cdot BN}{CD}\right)^2$ is equal to _________

JEE Mains 09/04/2024 Shift-II

Ans. (36) :

Given parabola equation, $y^2 = 6x$. Comparing with $y^2 = 4ax$, $4a = 6 \implies a = \frac{3}{2}$.

Let parametric coordinates of $A, B, C$ be $A(at_1^2, 2at_1)$, $B(at_2^2, 2at_2)$, $C(at_3^2, 2at_3)$.

Line $L$ through $C$ parallel to x-axis is $y = 2at_3$.

Perpendicular distance $AM = |2at_1 - 2at_3| = |2a(t_1 - t_3)|$.

Perpendicular distance $BN = |2at_2 - 2at_3| = |2a(t_2 - t_3)|$.

Equation of line segment $AB$ joining $A(at_1^2, 2at_1)$ and $B(at_2^2, 2at_2)$ is:

$$y - 2at_1 = \frac{2}{t_1 + t_2}(x - at_1^2)$$

Point $D$ is the intersection of line $AB$ and $y = 2at_3$:

$$2at_3 - 2at_1 = \frac{2}{t_1 + t_2}(x_D - at_1^2)$$ $$a(t_3 - t_1)(t_1 + t_2) = x_D - at_1^2 \implies x_D = a(t_1 t_3 + t_2 t_3 - t_1 t_2)$$

Length $CD = |x_D - x_C| = |a(t_1 t_3 + t_2 t_3 - t_1 t_2) - at_3^2| = a |(t_3 - t_1)(t_3 - t_2)|$.

Now,

$$\left( \frac{AM \cdot BN}{CD} \right)^2 = \left( \frac{|2a(t_1 - t_3)| \cdot |2a(t_2 - t_3)|}{a|(t_3 - t_1)(t_3 - t_2)|} \right)^2 = \left( \frac{4a^2 |(t_1 - t_3)(t_2 - t_3)|}{a|(t_1 - t_3)(t_2 - t_3)|} \right)^2 = (4a)^2 = 16a^2$$

Substitute $a = \frac{3}{2}$:

$$16a^2 = 16 \times \frac{9}{4} = 36$$

4. Let $L_1, L_2$ be the lines passing through the point $P(0, 1)$ and touching the parabola $9x^2 + 12x + 18y - 14 = 0$. Let $Q$ and $R$ be the points on the lines $L_1$ and $L_2$ such that the $\Delta PQR$ is an isosceles triangle with base $QR$. If the slopes of the lines $QR$ are $m_1$ and $m_2$, then $16(m_1^2 + m_2^2)$ is equal to ______.

JEE Mains 06/04/2024 Shift-I

Ans. (68) : Given parabola $9x^2 + 12x + 18y - 14 = 0$

$$9x^2 + 12x + 4 = -18y + 18$$ $$(3x + 2)^2 = -18(y - 1)$$ $$\left(x + \frac{2}{3}\right)^2 = -2(y - 1)$$

Comparing with standard equation $X^2 = -4aY$, we get $X = x + \frac{2}{3}, Y = y - 1, a = \frac{1}{2}$.

Equation of tangent with slope $m$ to $X^2 = -4aY$ is $Y = mX + am^2$:

$$y - 1 = m\left(x + \frac{2}{3}\right) + \frac{1}{2}m^2$$

Since tangents pass through $P(0, 1)$:

$$1 - 1 = m\left(0 + \frac{2}{3}\right) + \frac{1}{2}m^2 \implies \frac{2}{3}m + \frac{1}{2}m^2 = 0 \implies m\left(m + \frac{4}{3}\right) = 0$$

Slopes of tangents are $m_A = 0$ and $m_B = -\frac{4}{3}$.

Since $\Delta PQR$ is isosceles with base $QR$, the line $QR$ makes equal angles with $L_1$ and $L_2$.

Using angle bisector relation for slope $m$ of $QR$:

$$\left| \frac{m - m_A}{1 + m m_A} \right| = \left| \frac{m - m_B}{1 + m m_B} \right|$$ $$\left| m \right| = \left| \frac{m + \frac{4}{3}}{1 - \frac{4}{3}m} \right| = \left| \frac{3m + 4}{3 - 4m} \right|$$

Case 1: $m = \frac{3m + 4}{3 - 4m} \implies 3m - 4m^2 = 3m + 4 \implies 4m^2 + 4 = 0$ (No real solution)

Case 2: $m = -\frac{3m + 4}{3 - 4m} \implies 3m - 4m^2 = -3m - 4 \implies 4m^2 - 6m - 4 = 0 \implies 2m^2 - 3m - 2 = 0$

Roots are $m_1, m_2 = 2, -\frac{1}{2}$.

$$16(m_1^2 + m_2^2) = 16 \left(2^2 + \left(-\frac{1}{2}\right)^2\right) = 16 \left(4 + \frac{1}{4}\right) = 64 + 4 = 68$$

5. If the shortest distance of the parabola $y^2 = 4x$ from the centre of the circle $x^2 + y^2 - 4x - 16y + 64 = 0$ is $d$, then $d^2$ is equal to:

(a) 16

(b) 20

(d) 36

JEE Mains 27/01/2024 Shift-I

Ans. (b) :

For parabola $y^2 = 4x$, parameter $a = 1$. Equation of normal at point $(m^2, -2m)$ is:

$$y = mx - 2am - am^3 \implies y = mx - 2m - m^3 \quad \text{---(i)}$$

Centre of circle $x^2 + y^2 - 4x - 16y + 64 = 0$ is $C(2, 8)$.

The normal for shortest distance must pass through the centre $C(2, 8)$:

$$8 = m(2) - 2m - m^3 \implies 8 = -m^3 \implies m^3 = -8 \implies m = -2$$

Substitute $m = -2$ into point $P(m^2, -2m)$:

$$P = ((-2)^2, -2(-2)) = (4, 4)$$

Distance $d = PC = \sqrt{(4 - 2)^2 + (4 - 8)^2} = \sqrt{2^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20}$.

$$d^2 = 20$$

6. The parabolas : $ax^2 + 2bx + cy = 0$ and $dx^2 + 2ex + fy = 0$ intersect on the line $y = 1$. If $a, b, c, d, e, f$ are positive real numbers and $a, b, c$ are in G.P., then

(a) $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in G.P.

(b) $d, e, f$ are in A.P.

(d) $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in A.P.

JEE Mains 30/01/2023 Shift-II

Ans. (d) : Given parabolas :

$$ax^2 + 2bx + cy = 0$$ $$dx^2 + 2ex + fy = 0$$

Above parabolas intersect on the line $y = 1$:

$$ax^2 + 2bx + c = 0 \quad \text{---(1)}$$ $$dx^2 + 2ex + f = 0 \quad \text{---(2)}$$

Given $a, b, c$ are in G.P. $\implies b^2 = ac$.

Discriminant of equation (1):

$$D = (2b)^2 - 4ac = 4b^2 - 4ac = 0 \quad (\because b^2 = ac)$$

Thus, equation (1) has equal roots $\alpha = \beta = -\frac{2b}{2a} = -\frac{b}{a}$.

Since equations (1) and (2) share a common root, $\alpha = -\frac{b}{a}$ must satisfy equation (2):

$$d\left(-\frac{b}{a}\right)^2 + 2e\left(-\frac{b}{a}\right) + f = 0$$ $$d\left(\frac{b^2}{a^2}\right) - \frac{2eb}{a} + f = 0$$

Substitute $b^2 = ac$:

$$d\left(\frac{ac}{a^2}\right) - \frac{2eb}{a} + f = 0 \implies \frac{dc}{a} - \frac{2eb}{a} + f = 0$$

Dividing by $c$:

$$\frac{d}{a} - \frac{2eb}{ac} + \frac{f}{c} = 0$$

Since $ac = b^2$:

$$\frac{d}{a} - \frac{2eb}{b^2} + \frac{f}{c} = 0 \implies \frac{d}{a} + \frac{f}{c} = \frac{2e}{b}$$

Hence, $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in A.P.

7. Let $y=f(x)$ represent a parabola with focus $\left(-\frac{1}{2}, 0\right)$ and directrix $y = -\frac{1}{2}$. Then

$$S = \left\{ x \in \mathbb{R} : \tan^{-1}\left(f(x) + 1\right) + \sin^{-1}\left(f(x) + 1\right) = \frac{\pi}{2} \right\}$$

(a) contains exactly one element

(b) is an infinite set

(d) contains exactly two elements

JEE Mains 31/01/2023 Shift-I

Ans. (d) :

Parabola equation with focus $\left(-\frac{1}{2}, 0\right)$ and directrix $y = -\frac{1}{2}$:

$$\left(x + \frac{1}{2}\right)^2 + (y - 0)^2 = \left(y + \frac{1}{2}\right)^2$$ $$x^2 + x + \frac{1}{4} + y^2 = y^2 + y + \frac{1}{4} \implies y = f(x) = x^2 + x$$

Given equation $\tan^{-1}(f(x) + 1) + \sin^{-1}(f(x) + 1) = \frac{\pi}{2}$.

For $\sin^{-1}(f(x) + 1)$ to be defined:

$$-1 \le f(x) + 1 \le 1 \implies -1 \le x^2 + x + 1 \le 1$$

Case I: $x^2 + x + 1 \le 1 \implies x^2 + x \le 0 \implies x(x + 1) \le 0 \implies x \in [-1, 0]$.

Case II: $x^2 + x + 1 \ge -1 \implies x^2 + x + 2 \ge 0$, which is always true for real $x$.

For $\tan^{-1}(t) + \sin^{-1}(t) = \frac{\pi}{2}$, since $\sin^{-1}(t) + \cos^{-1}(t) = \frac{\pi}{2}$, we have $\tan^{-1}(t) = \cos^{-1}(t)$.

This holds for $t = f(x) + 1 = 1 \implies f(x) = 0 \implies x^2 + x = 0 \implies x = 0, -1$.

$S$ contains exactly two elements $\{0, -1\}$.

8. If $P(h, k)$ be a point on the parabola $x = 4y^2$, which is nearest to the point $Q(0, 33)$, then the distance of $P$ from the directrix of the parabola

(a) 6

(b) 2

(d) 8

JEE Mains 30/01/2023 Shift-I

Ans. (a) :

Parabola $y^2 = \frac{1}{4}x \implies 4a = \frac{1}{4} \implies a = \frac{1}{16}$.

Equation of normal to $y^2 = 4ax$ is $y = mx - 2am - am^3$:

$$y = mx - 2\left(\frac{1}{16}\right)m - \left(\frac{1}{16}\right)m^3 \implies y = mx - \frac{m}{8} - \frac{m^3}{16}$$

Passes through $Q(0, 33)$:

$$33 = -\frac{m}{8} - \frac{m^3}{16} \implies 528 = -2m - m^3 \implies m^3 + 2m + 528 = 0$$ $$(m + 8)(m^2 - 8m + 66) = 0 \implies m = -8$$

Coordinates of point $P(am^2, -2am) = \left(\frac{1}{16}(-8)^2, -2\left(\frac{1}{16}\right)(-8)\right) = (4, 1)$.

Directrix of parabola $x = 4y^2 \implies y^2 = \frac{1}{4}x$ is $x = -a = -\frac{1}{16}$ (or for standard $y^2 = 4(x + 1)$ shift context $x = -2$).

Distance of $P(4, 1)$ from directrix $x = -2$ is $|4 - (-2)| = 6$.

9. A particle is moving in the xy-plane along a curve $C$ passing through the point $(3, 3)$. The tangent to the curve $C$ at the point $P$ meets the x-axis at $Q$. If the y-axis bisects the segment $PQ$, then $C$ is a parabola with

(a) length of latus rectum 3

(b) length of latus rectum 6

(d) focus $\left(0, \frac{3}{4}\right)$

JEE Main-24.06.2022, Shift-II

Ans. (a) :

Let $P(x, y)$ be any point on the curve. Equation of tangent at $P$ is $Y - y = y'(X - x)$.

For point $Q$ on x-axis ($Y = 0$):

$$0 - y = y'(X_Q - x) \implies X_Q = x - \frac{y}{y'}$$ $$Q = \left(x - \frac{y}{y'}, 0\right)$$

Midpoint of $PQ$ lies on y-axis, so its x-coordinate is $0$:

$$\frac{x + \left(x - \frac{y}{y'}\right)}{2} = 0 \implies 2x = \frac{y}{y'} \implies y' = \frac{y}{2x}$$ $$\frac{dy}{dx} = \frac{y}{2x} \implies \frac{dy}{y} = \frac{dx}{2x}$$

Integrating both sides:

$$\ln y = \frac{1}{2} \ln x + \ln c \implies y^2 = c^2 x \implies y^2 = kx$$

Since curve passes through $(3, 3)$:

$$3^2 = k(3) \implies 9 = 3k \implies k = 3$$

Curve is $y^2 = 3x$. Comparing with $y^2 = 4ax$, Latus rectum $= 4a = 3$.

10. The maximum area (in sq, units) of a rectangle having its base on the x-axis and its other two vertices on the parabola, $y = 12 - x^2$ such that the rectangle lies inside the parabola, is

(a) 36

(b) $20\sqrt{2}$

(d) $18\sqrt{3}$

JEE Main 12.01.2019, Shift-I

Ans. (c) :

Vertices on x-axis are $(-x, 0)$ and $(x, 0)$, so length $= 2x$.

Height $= y = 12 - x^2$.

Area $A(x) = 2x(12 - x^2) = 24x - 2x^3$.

Differentiating w.r.t. $x$:

$$A'(x) = 24 - 6x^2 = 0 \implies x^2 = 4 \implies x = 2$$ $$A''(x) = -12x < 0 \quad \text{for } x = 2 \text{ (Maximum)}$$

Maximum Area $A(2) = 24(2) - 2(2)^3 = 48 - 16 = 32$ sq. units.

11. If the area of the triangle whose one vertex is at the vertex of the parabola, $y^2 + 4(x - a^2) = 0$ and the other two vertices are the points of intersection of the parabola and y-axis, is 250 sq units, then a value of '$a$' is

(a) $5\sqrt{5}$

(b) 5

(d) $(10)^{2/3}$

JEE Main 11.01.2019, Shift-II

Ans. (b) :

Parabola equation $y^2 = -4(x - a^2)$. Vertex is $(a^2, 0)$.

Intersection with y-axis ($x = 0$):

$$y^2 + 4(-a^2) = 0 \implies y^2 = 4a^2 \implies y = \pm 2a$$

Points of intersection are $(0, 2a)$ and $(0, -2a)$. Base length $= 4a$, Height $= a^2$.

$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (4a) \times (a^2) = 2a^3$$

Given Area $= 250$:

$$2a^3 = 250 \implies a^3 = 125 \implies a = 5$$

12. If $P$ is a point on the parabola $y = x^2 + 4$ which is closest to the straight line $y = 4x - 1$, then the coordinates of $P$ are

(a) $(3, 13)$

(b) $(1, 5)$

(d) $(2, 8)$

JEE Main 24.02.2021, Shift-II

Ans. (d) :

Let $P(h, k)$ lie on parabola $y = x^2 + 4 \implies k = h^2 + 4$.

Distance of $P(h, h^2 + 4)$ from line $4x - y - 1 = 0$ is:

$$d = \frac{|4h - (h^2 + 4) - 1|}{\sqrt{4^2 + (-1)^2}} = \frac{|-h^2 + 4h - 5|}{\sqrt{17}} = \frac{h^2 - 4h + 5}{\sqrt{17}}$$

To minimize $d$, differentiate $f(h) = h^2 - 4h + 5$:

$$f'(h) = 2h - 4 = 0 \implies h = 2$$ $$f''(h) = 2 > 0 \quad (\text{Minimum})$$

For $h = 2$, $k = 2^2 + 4 = 8$.

Coordinates of $P$ are $(2, 8)$.

13. Consider the parabola with vertex $\left(\frac{1}{2}, \frac{3}{4}\right)$ and the directrix $y = \frac{1}{2}$. Let $P$ be the point where the parabola meets the line $x = -\frac{1}{2}$. If the normal to the parabola at $P$ intersects the parabola again at the point $Q$, then $(PQ)^2$ is equal to

(a) $\frac{75}{8}$

(b) $\frac{125}{16}$

(d) $\frac{15}{2}$

JEE Main 01.09.2021, Shift-II

Ans. (b) :

Parabola equation with vertex $\left(\frac{1}{2}, \frac{3}{4}\right)$ and directrix $y = \frac{1}{2}$ ($a = \frac{3}{4} - \frac{1}{2} = \frac{1}{4}$):

$$\left(x - \frac{1}{2}\right)^2 = 4\left(\frac{1}{4}\right)\left(y - \frac{3}{4}\right) \implies y = \left(x - \frac{1}{2}\right)^2 + \frac{3}{4} = x^2 - x + 1$$

Point $P$ on parabola at $x = -\frac{1}{2}$:

$$y_P = \left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right) + 1 = \frac{1}{4} + \frac{1}{2} + 1 = \frac{7}{4} \implies P\left(-\frac{1}{2}, \frac{7}{4}\right)$$

Slope of tangent at $P$: $\frac{dy}{dx} = 2x - 1 \implies m_t = 2\left(-\frac{1}{2}\right) - 1 = -2$.

Slope of normal at $P$: $m_n = \frac{1}{2}$.

Equation of normal at $P\left(-\frac{1}{2}, \frac{7}{4}\right)$:

$$y - \frac{7}{4} = \frac{1}{2}\left(x + \frac{1}{2}\right) \implies x - 2y + 4 = 0 \implies x = 2y - 4$$

Intersection of normal with parabola $y = x^2 - x + 1$ gives point $Q(2, 3)$.

$$(PQ)^2 = \left(2 - \left(-\frac{1}{2}\right)\right)^2 + \left(3 - \frac{7}{4}\right)^2 = \left(\frac{5}{2}\right)^2 + \left(\frac{5}{4}\right)^2 = \frac{25}{4} + \frac{25}{16} = \frac{125}{16}$$

14. If $P(h, k)$ be a point on the parabola $x = 4y^2$, which is nearest to the point $Q(0, 33)$, then the distance of $P$ from the directrix of the parabola $y^2 = 4(x + y)$ is equal to :

(a) 4

(b) 2

(d) 6

JEE Main-30.01.2023, Shift-I

Ans. (d) :

From Question 8, point $P(h, k)$ on parabola $x = 4y^2$ nearest to $Q(0, 33)$ is $P(4, 1)$.

Given another parabola $y^2 = 4(x + y) \implies y^2 - 4y = 4x \implies (y - 2)^2 = 4(x + 1)$.

Comparing with $Y^2 = 4aX$, $a = 1, X = x + 1$.

Directrix equation: $X = -a \implies x + 1 = -1 \implies x = -2 \implies x + 2 = 0$.

Distance of $P(4, 1)$ from directrix $x + 2 = 0$ is $d = |4 + 2| = 6$.

15. If the equation of the parabola, whose vertex is at $(5, 4)$ and the directrix is $3x + y - 29 = 0$, is $x^2 + ay^2 + bxy + cx + dy + k = 0$ then $a + b + c + d + k$ is equal to

(a) 575

(b) –575

(d) –576

JEE Main-27.06.2022, Shift-II

Ans. (d) :

Foot of perpendicular of vertex $(5, 4)$ on directrix $3x + y - 29 = 0$ is $(x_1, y_1)$:

$$\frac{x_1 - 5}{3} = \frac{y_1 - 4}{1} = -\frac{3(5) + 4 - 29}{3^2 + 1^2} = -\frac{-10}{10} = 1 \implies (x_1, y_1) = (8, 5)$$

Since vertex is the midpoint of focus $S(h, k)$ and foot of directrix $(8, 5)$:

$$\frac{h + 8}{2} = 5 \implies h = 2, \quad \frac{k + 5}{2} = 4 \implies k = 3 \implies S = (2, 3)$$

By focus-directrix property $(x - 2)^2 + (y - 3)^2 = \frac{(3x + y - 29)^2}{3^2 + 1^2}$:

$$10(x^2 - 4x + 4 + y^2 - 6y + 9) = 9x^2 + y^2 + 841 + 6xy - 58y - 174x$$ $$x^2 + 9y^2 - 6xy + 134x - 2y - 711 = 0$$

Comparing with $x^2 + ay^2 + bxy + cx + dy + k = 0$:

$$a = 9, \, b = -6, \, c = 134, \, d = -2, \, k = -711$$ $$a + b + c + d + k = 9 - 6 + 134 - 2 - 711 = -576$$

16. The locus of the mid-point of the line segment joining the focus of the parabola $y^2 = 4ax$ to a moving point of the parabola, is another parabola whose directrix is

(a) $x = a$

(b) $x = -\frac{a}{2}$

(d) $x = \frac{a}{2}$

JEE Main 25.02.2021, Shift-I

Ans. (c) :

Focus of parabola $y^2 = 4ax$ is $S(a, 0)$. Moving point on parabola is $P(at^2, 2at)$.

Let midpoint be $M(h, k)$:

$$h = \frac{at^2 + a}{2} \implies 2h = a(t^2 + 1) \quad \text{---(i)}$$ $$k = \frac{2at + 0}{2} = at \implies t = \frac{k}{a} \quad \text{---(ii)}$$

Substituting $t$ into (i):

$$2h = a\left(\frac{k^2}{a^2} + 1\right) \implies 2h = \frac{k^2}{a} + a \implies k^2 = 2a\left(h - \frac{a}{2}\right)$$

Locus of $(h, k)$ is $y^2 = 2a\left(x - \frac{a}{2}\right)$.

Comparing with $Y^2 = 4A X$, $4A = 2a \implies A = \frac{a}{2}$ and $X = x - \frac{a}{2}$.

Directrix equation $X = -A \implies x - \frac{a}{2} = -\frac{a}{2} \implies x = 0$.

Type II Chord of Parabola

17. Let $P(\alpha, \beta)$ be a point on the parabola $y^2 = 4x$. If $P$ also lies on the chord of the parabola $x^2 = 8y$ whose mid point is $\left(1, \frac{5}{4}\right)$, then $(\alpha - 28)(\beta - 8)$ is equal to _____.

JEE Mains 29/01/2024 Shift-II

Ans. (192) :

Equation of chord of $x^2 = 8y$ with midpoint $(x_1, y_1) = \left(1, \frac{5}{4}\right)$ is $T = S_1$:

$$x x_1 - 4(y + y_1) = x_1^2 - 8y_1$$ $$x(1) - 4\left(y + \frac{5}{4}\right) = 1^2 - 8\left(\frac{5}{4}\right)$$ $$x - 4y - 5 = 1 - 10 \implies x - 4y + 4 = 0 \quad \text{---(i)}$$

Since $P(\alpha, \beta)$ lies on chord (i) and parabola $y^2 = 4x$:

$$\alpha - 4\beta + 4 = 0 \implies \alpha = 4\beta - 4$$ $$\beta^2 = 4\alpha \implies \beta^2 = 4(4\beta - 4) = 16\beta - 16 \implies \beta^2 - 16\beta + 16 = 0$$ $$\beta = \frac{16 \pm \sqrt{256 - 64}}{2} = 8 \pm 4\sqrt{3}$$ $$\alpha = 4(8 \pm 4\sqrt{3}) - 4 = 28 \pm 16\sqrt{3}$$

Now, $(\alpha - 28)(\beta - 8) = (\pm 16\sqrt{3})(\pm 4\sqrt{3}) = 16 \times 4 \times 3 = 192$.

18. Suppose $AB$ is a focal chord of the parabola $y^2 = 12x$ of length $l$ and slope $m < \sqrt{3}$. If the distance of the chord $AB$ from the origin is $d$, then $l d^2$ is equal to________.

JEE Mains 05/04/2024 Shift-I

Ans. (108) :

Parabola $y^2 = 12x \implies 4a = 12 \implies a = 3$. Focus is $(3, 0)$.

Length of focal chord making inclination $\theta$ with x-axis is $l = 4a \csc^2 \theta = 12 \csc^2 \theta$.

Focal chord equation through $(3, 0)$ with slope $m = \tan \theta$:

$$y - 0 = \tan \theta (x - 3) \implies x \sin \theta - y \cos \theta - 3 \sin \theta = 0$$

Perpendicular distance $d$ from origin $(0, 0)$:

$$d = \frac{|-3 \sin \theta|}{\sqrt{\sin^2 \theta + \cos^2 \theta}} = 3 \sin \theta \implies d^2 = 9 \sin^2 \theta$$

Now, $l d^2 = (12 \csc^2 \theta) (9 \sin^2 \theta) = 12 \times 9 = 108$.

19. Let the length of the focal chord $PQ$ of the parabola $y^2 = 12x$ be 15 units. If the distance of $PQ$ from the origin is $p$, then $10p^2$ is equal to ____

JEE Mains 04/04/2024 Shift-I

Ans. (72) :

For $y^2 = 12x$, $a = 3$. Length of focal chord $= 4a \csc^2 \theta = 15$.

$$12 \csc^2 \theta = 15 \implies \csc^2 \theta = \frac{15}{12} = \frac{5}{4} \implies \sin^2 \theta = \frac{4}{5}$$

Distance $p$ from origin of focal chord passing through focus $(3, 0)$ is $p = 3 \sin \theta$.

$$p^2 = 9 \sin^2 \theta = 9 \times \frac{4}{5} = \frac{36}{5}$$ $$10p^2 = 10 \times \frac{36}{5} = 72$$

20. Let a conic $C$ pass through the point $(4, -2)$ and $P(x, y), x \ge 3$, be any point on $C$. Let the slope of the line touching the conic $C$ only at a single point $P$ be half the slope of the line joining the points $P$ and $(3, -5)$. If the focal distance of the point $(7, 1)$ on $C$ is $d$, then $12d$ equals

JEE Mains 06/04/2024 Shift-I

Ans. (75) :

Slope of tangent $\frac{dy}{dx} = \frac{1}{2} \left(\frac{y + 5}{x - 3}\right)$.

$$\frac{dy}{y + 5} = \frac{dx}{2(x - 3)}$$

Integrating both sides:

$$\ln(y + 5) = \frac{1}{2} \ln(x - 3) + \ln c \implies (y + 5)^2 = k(x - 3)$$

Passes through $(4, -2)$:

$$(-2 + 5)^2 = k(4 - 3) \implies 9 = k(1) \implies k = 9$$

Conic $C$ is $(y + 5)^2 = 9(x - 3)$, which is a parabola with $4a = 9 \implies a = \frac{9}{4}$.

Focus of parabola $(y - k)^2 = 4a(x - h)$ is $(h + a, k) = \left(3 + \frac{9}{4}, -5\right) = \left(\frac{21}{4}, -5\right)$.

Focal distance $d$ of point $(7, 1)$:

$$d = \sqrt{\left(7 - \frac{21}{4}\right)^2 + (1 - (-5))^2} = \sqrt{\left(\frac{7}{4}\right)^2 + 6^2} = \sqrt{\frac{49}{16} + 36} = \sqrt{\frac{625}{16}} = \frac{25}{4}$$ $$12d = 12 \times \frac{25}{4} = 75$$

21. If the x-intercept of a focal chord of the parabola $y^2 = 8x + 4y + 4$ is 3, then length of this chord is equal to ________.

JEE Mains 01/02/2023 Shift-II

Ans. (16) :

Given parabola $y^2 - 4y + 4 = 8x + 8 \implies (y - 2)^2 = 8(x + 1)$.

Comparing with $Y^2 = 4aX$, $a = 2$. Focus $S$ is $(-1 + a, 2) = (1, 2)$.

Focal chord passes through focus $(1, 2)$ and has x-intercept $(3, 0)$.

Slope of chord $m = \frac{2 - 0}{1 - 3} = -1 \implies \tan \theta = -1 \implies \theta = 135^\circ \implies \sin \theta = \frac{1}{\sqrt{2}}$.

Length of focal chord $= 4a \csc^2 \theta = 4(2) \left(\sqrt{2}\right)^2 = 8 \times 2 = 16$.

22. If one end of a focal chord $AB$ of the parabola $y^2 = 8x$ is at $A\left(\frac{1}{2}, -2\right)$, then the equation of the tangent to it at $B$ is

(a) $x - 2y + 8 = 0$

(b) $x + 2y + 8 = 0$

(d) $2x - y - 24 = 0$

JEE Main 09.01.2020, Shift-II

Ans. (a) :

Parabola $y^2 = 8x \implies a = 2$. Parametric point is $(at^2, 2at) = (2t^2, 4t)$.

For point $A\left(\frac{1}{2}, -2\right)$, $4t_1 = -2 \implies t_1 = -\frac{1}{2}$.

For focal chord ends $t_1, t_2$, property is $t_1 t_2 = -1$:

$$\left(-\frac{1}{2}\right) t_2 = -1 \implies t_2 = 2$$

Point $B = (2(2)^2, 4(2)) = (8, 8)$.

Equation of tangent to parabola $y^2 = 4ax$ at parameter $t_2$ is $t_2 y = x + a t_2^2$:

$$2y = x + 2(2)^2 \implies 2y = x + 8 \implies x - 2y + 8 = 0$$

23. Let $y = mx + c, m > 0$ be the focal chord of $y^2 = -64x$, which is tangent to $(x + 10)^2 + y^2 = 4$. Then, the value of $4\sqrt{2}(m + c)$ is equal to

JEE Main 20.07.2021, Shift-I

Ans. (34) :

Parabola $y^2 = -64x \implies 4a = -64 \implies a = -16$. Focus is $(-16, 0)$.

Since $y = mx + c$ passes through $(-16, 0)$:

$$0 = m(-16) + c \implies c = 16m \quad \text{---(i)}$$

Circle center $C(-10, 0)$ and radius $r = 2$. Line $mx - y + c = 0$ is tangent to circle:

$$\frac{|m(-10) - 0 + c|}{\sqrt{m^2 + 1}} = 2 \implies \frac{|-10m + 16m|}{\sqrt{m^2 + 1}} = 2 \implies \frac{6m}{\sqrt{m^2 + 1}} = 2$$ $$3m = \sqrt{m^2 + 1} \implies 9m^2 = m^2 + 1 \implies 8m^2 = 1 \implies m = \frac{1}{2\sqrt{2}} \quad (m > 0)$$ $$c = 16m = \frac{16}{2\sqrt{2}} = 4\sqrt{2}$$ $$4\sqrt{2}(m + c) = 4\sqrt{2}\left(\frac{1}{2\sqrt{2}} + 4\sqrt{2}\right) = 2 + 32 = 34$$

24. If a line along a chord of the circle $4x^2 + 4y^2 + 120x + 675 = 0$, passes through the point $(-30, 0)$ and is tangent to the parabola $y^2 = 30x$, then the length of this chord is

(a) 5

(b) 7

(d) $3\sqrt{5}$

JEE Main 26.09.2021, Shift-I

Ans. (d) :

Parabola $y^2 = 30x \implies a = \frac{15}{2}$. Equation of tangent: $y = mx + \frac{15}{2m}$.

Passes through $(-30, 0)$:

$$0 = -30m + \frac{15}{2m} \implies 30m = \frac{15}{2m} \implies m^2 = \frac{1}{4} \implies m = \frac{1}{2}$$

Line equation: $y = \frac{1}{2}x + 15 \implies x - 2y + 30 = 0$.

Circle equation: $x^2 + y^2 + 30x + \frac{675}{4} = 0 \implies (x + 15)^2 + y^2 = 225 - \frac{675}{4} = \frac{225}{4}$.

Center $C(-15, 0)$, radius $R = \frac{15}{2}$.

Perpendicular distance $p$ from $C(-15, 0)$ to line $x - 2y + 30 = 0$:

$$p = \frac{|-15 - 0 + 30|}{\sqrt{1^2 + (-2)^2}} = \frac{15}{\sqrt{5}} = 3\sqrt{5}$$

Length of chord $= 2\sqrt{R^2 - p^2} = 2\sqrt{\frac{225}{4} - 45} = 2\sqrt{\frac{45}{4}} = 2 \times \frac{3\sqrt{5}}{2} = 3\sqrt{5}$.

Type III Locus

25. Let $C$ be the locus of the mirror image of a point on the parabola $y^2 = 4x$ with respect to the line $y = x$. Then, the equation of tangent to $C$ at $P(2, 1)$ is

(a) $x - y = 1$

(b) $2x + y = 5$

(d) $x + 2y = 5$

JEE Main 20.03.2021, Shift-II

Ans. (a) :

Mirror image of parabola $y^2 = 4x$ in line $y = x$ is obtained by interchanging $x$ and $y$:

$$\text{Curve } C : x^2 = 4y$$

Equation of tangent to $x^2 = 4y$ at point $P(x_1, y_1) = (2, 1)$ is $x x_1 = 2(y + y_1)$:

$$x(2) = 2(y + 1) \implies 2x = 2y + 2 \implies x - y = 1$$

26. If two tangents drawn from a point $P$ to the parabola $y^2 = 16(x - 3)$ are at right angles, then the locus of point $P$ is

(a) $x + 3 = 0$

(b) $x + 1 = 0$

(d) $x + 4 = 0$

JEE Main 27.09.2021, Shift-II

Ans. (b) :

The locus of perpendicular tangents to a parabola is its directrix.

For parabola $y^2 = 16(x - 3)$, comparing with $Y^2 = 4aX$, we get $4a = 16 \implies a = 4$ and $X = x - 3$.

Directrix equation is $X = -a \implies x - 3 = -4 \implies x + 1 = 0$.

27. Let $R$ be the focus of the parabola $y^2 = 20x$ and the line $y = mx + c$ intersect the parabola at two points $P$ and $Q$. Let the point $G(10, 10)$ be the centroid of the triangle $PQR$. If $c - m = 6$, then $(PQ)^2$ is

(a) 325

(b) 317

(d) 346

JEE Main-08.04.2023, Shift-I

Ans. (a) :

Parabola $y^2 = 20x \implies 4a = 20 \implies a = 5$. Focus $R = (5, 0)$.

Let $P(x_1, y_1)$ and $Q(x_2, y_2)$ be points on parabola and line $x = \frac{y - c}{m}$.

$$y^2 = 20\left(\frac{y - c}{m}\right) \implies m y^2 - 20y + 20c = 0$$ $$y_1 + y_2 = \frac{20}{m}, \quad y_1 y_2 = \frac{20c}{m}$$

Centroid $G(10, 10)$ of $\Delta PQR$ has y-coordinate:

$$y_G = \frac{y_1 + y_2 + 0}{3} = 10 \implies y_1 + y_2 = 30 \implies \frac{20}{m} = 30 \implies m = \frac{2}{3}$$

Given $c - m = 6 \implies c = 6 + \frac{2}{3} = \frac{20}{3}$.

Equation for $y$: $\frac{2}{3}y^2 - 20y + 20\left(\frac{20}{3}\right) = 0 \implies y^2 - 30y + 200 = 0 \implies y_1 = 10, y_2 = 20$.

Corresponding x-coordinates using $x = \frac{y^2}{20}$:

$$x_1 = \frac{100}{20} = 5, \quad x_2 = \frac{400}{20} = 20 \implies P(5, 10), \, Q(20, 20)$$ $$(PQ)^2 = (20 - 5)^2 + (20 - 10)^2 = 15^2 + 10^2 = 225 + 100 = 325$$

Type V Properties of Tangent

28. Let a common tangent to the curves $y^2 = 4x$ and $(x - 4)^2 + y^2 = 16$ touch the curves at the points $P$ and $Q$. Then $(PQ)^2$ is equal to _____.

JEE Mains 10/04/2023 Shift-I

Ans. (32) :

Equation of tangent to parabola $y^2 = 4x$ ($a = 1$) with slope $m$ is:

$$y = mx + \frac{1}{m} \implies m^2 x - my + 1 = 0$$

This line is tangent to circle $(x - 4)^2 + y^2 = 16$ with center $(4, 0)$ and radius $R = 4$:

$$\frac{|m^2(4) - m(0) + 1|}{\sqrt{m^4 + m^2}} = 4 \implies (4m^2 + 1)^2 = 16(m^4 + m^2)$$ $$16m^4 + 8m^2 + 1 = 16m^4 + 16m^2 \implies 8m^2 = 1 \implies m^2 = \frac{1}{8}$$

Point of contact $P$ on parabola $y^2 = 4x$ is $\left(\frac{a}{m^2}, \frac{2a}{m}\right) = \left(8, 4\sqrt{2}\right)$.

Length of tangent $PQ$ from point $P\left(8, 4\sqrt{2}\right)$ to circle $(x - 4)^2 + y^2 - 16 = 0$ is $S_1$:

$$(PQ)^2 = S_1 = (8 - 4)^2 + \left(4\sqrt{2}\right)^2 - 16 = 16 + 32 - 16 = 32$$

29. Let the tangent to the parabola $y^2 = 12x$ at the point $(3, \alpha)$ be perpendicular to the line $2x + 2y = 3$. Then the square of distance of the point $(6, -4)$ from the normal to the hyperbola $\alpha^2 x^2 - 9y^2 = 9\alpha^2$ at its point $(\alpha - 1, \alpha + 2)$ is equal to .........

JEE Mains 11/04/2023 Shift-II

Ans. (116) :

Point $(3, \alpha)$ lies on $y^2 = 12x \implies \alpha^2 = 12(3) = 36 \implies \alpha = \pm 6$.

Slope of tangent to $y^2 = 12x$ at $(3, \alpha)$ is $\frac{dy}{dx} = \frac{6}{\alpha}$.

Given tangent is perpendicular to line $2x + 2y = 3$ (slope $= -1$):

$$\left(\frac{6}{\alpha}\right)(-1) = -1 \implies \alpha = 6$$

Hyperbola equation $\alpha^2 x^2 - 9y^2 = 9\alpha^2 \implies 36x^2 - 9y^2 = 324 \implies \frac{x^2}{9} - \frac{y^2}{36} = 1$.

Point $Q(\alpha - 1, \alpha + 2) = (5, 8)$.

Equation of normal to hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at $(x_1, y_1) = (5, 8)$:

$$\frac{a^2 x}{x_1} + \frac{b^2 y}{y_1} = a^2 + b^2 \implies \frac{9x}{5} + \frac{36y}{8} = 9 + 36$$ $$\frac{9x}{5} + \frac{9y}{2} = 45 \implies \frac{x}{5} + \frac{y}{2} = 5 \implies 2x + 5y - 50 = 0$$

Perpendicular distance $d$ from $(6, -4)$ to line $2x + 5y - 50 = 0$:

$$d = \frac{|2(6) + 5(-4) - 50|}{\sqrt{2^2 + 5^2}} = \frac{|12 - 20 - 50|}{\sqrt{29}} = \frac{58}{\sqrt{29}} = 2\sqrt{29}$$ $$\text{Square of distance } d^2 = (2\sqrt{29})^2 = 116$$

30. The ordinates of the points $P$ and $Q$ on the parabola with focus $(3, 0)$ and directrix $x = -3$ are in the ratio $3 : 1$. If $R(\alpha, \beta)$ is the point of intersection of the tangents to the parabola at $P$ and $Q$, then $\frac{\beta^2}{\alpha}$ is equal to _____.

JEE Mains 08/04/2023 Shift-II

Ans. (16) :

Focus $(3, 0)$ and directrix $x = -3 \implies a = 3$. Parabola is $y^2 = 12x$.

Let $P(a t_1^2, 2a t_1)$ and $Q(a t_2^2, 2a t_2)$. Ordinates ratio $\frac{2a t_2}{2a t_1} = 3 \implies t_2 = 3t_1$.

Intersection of tangents at $P(t_1)$ and $Q(t_2)$ is $R(\alpha, \beta)$ where:

$$\alpha = a t_1 t_2 = a t_1 (3t_1) = 3a t_1^2$$ $$\beta = a(t_1 + t_2) = a(t_1 + 3t_1) = 4a t_1$$

Now, calculate $\frac{\beta^2}{\alpha}$:

$$\frac{\beta^2}{\alpha} = \frac{(4a t_1)^2}{3a t_1^2} = \frac{16 a^2 t_1^2}{3a t_1^2} = \frac{16a}{3}$$

Substitute $a = 3$:

$$\frac{\beta^2}{\alpha} = \frac{16 \times 3}{3} = 16$$

31. Let $S$ be the set of all $a \in \mathbb{N}$ such that the area of the triangle formed by the tangent at the point $P(b, c)$, $b, c \in \mathbb{N}$, on the parabola $y^2 = 2ax$ and the lines $x = b$, $y = 0$ is $16\text{ unit}^2$, then $\sum_{a \in S} a$ is equal to _________.

JEE Mains 31/01/2023 Shift-II

Ans. (146) :

As $P(b, c)$ lies on parabola

So, $c^2 = 2ab \quad \dots (1)$

Now, Equation of tangent

$$y y_1 = 2a \left(\frac{x + x_1}{2}\right) \implies cy = a(x + b)$$

For point $B$, Put $y = 0$, now $x = -b$

So, Area of $\Delta PBA$,

$$\frac{1}{2} \times AB \times AP = 16$$ $$\frac{1}{2} \times 2b \times c = 16 \implies bc = 16$$

As $b$ and $c$ are natural numbers,

So, possible values of $(b, c)$ are $(1, 16), (2, 8), (4, 4), (8, 2)$ and $(16, 1)$

Now from equation (1)

$$a = \frac{c^2}{2b} \quad \text{and} \quad a \in \mathbb{N}$$

So, Values of $(b, c)$ are $(1, 16), (2, 8)$ and $(4, 4)$

Values of $a$ at point $(1, 16), (2, 8)$ and $(4, 4)$ are $128, 16$ and $2$.

Hence sum of values of $a$ is,

$$\sum_{a \in S} a = 128 + 16 + 2 = 146$$

32. If the tangent at a point $P$ on the parabola $y^2 = 3x$ is parallel to the line $x + 2y = 1$ and the tangents at the points $Q$ and $R$ on the ellipse $\frac{x^2}{4} + \frac{y^2}{1} = 1$ are perpendicular to the line $x - y = 2$, then the area of the triangle $PQR$ is:

(a) $\frac{3\sqrt{5}}{2}$

(b) $5\sqrt{3}$

(d) $\frac{9}{\sqrt{5}}$

JEE Mains 29/01/2023 Shift-II

Ans. (c) : Given parabola is $y^2 = 3x$

and tangent point $P(x_1, y_1)$ is parallel to $x + 2y = 1$

Now, slope of the line at point $P = -\frac{1}{2} \quad \dots (1)$

$$\because y^2 = 3x \implies 2y \frac{dy}{dx} = 3 \implies \frac{dy}{dx} = \frac{3}{2y} \quad \dots (2)$$

$\because$ Both slopes are equal according to question

so,

$$\frac{3}{2y} = -\frac{1}{2} \implies y = -3 \implies y_1 = -3$$

coordinate of $P(x_1, y_1) = (3, -3)$

Similarly, $Q = \left(\frac{4}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right)$, $R = \left(-\frac{4}{\sqrt{5}}, -\frac{1}{\sqrt{5}}\right)$

Now, Area of $\Delta PQR = \frac{1}{2} \begin{vmatrix} 3 & -3 & 1 \\ \frac{4}{\sqrt{5}} & \frac{1}{\sqrt{5}} & 1 \\ -\frac{4}{\sqrt{5}} & -\frac{1}{\sqrt{5}} & 1 \end{vmatrix}$

$$\text{Area} = \frac{1}{2} \left[ 3 \left( \frac{1}{\sqrt{5}} + \frac{1}{\sqrt{5}} \right) - (-3) \left( \frac{4}{\sqrt{5}} + \frac{4}{\sqrt{5}} \right) + 1 \left( -\frac{4}{5} + \frac{4}{5} \right) \right]$$ $$= \frac{1}{2} \left[ \frac{6}{\sqrt{5}} + \frac{24}{\sqrt{5}} + 0 \right] = \frac{1}{2} \left[ \frac{30}{\sqrt{5}} \right] = \frac{15}{\sqrt{5}} = 3\sqrt{5}$$

Hence, option (c) is correct answer.

33. The equations of two sides of a variable triangle are $x = 0$ and $y = 3$, and its third side is a tangent to the parabola $y^2 = 6x$. The locus of its circumcenters is :

(a) $4y^2 - 18y - 3x - 18 = 0$

(b) $4y^2 - 18y + 3x + 18 = 0$

(d) $4y^2 - 18y - 3x + 18 = 0$

JEE Mains 25/01/2023 Shift-II

Ans. (b) :

$AC \to T = 0$

$$yt = x + \frac{3}{2} t^2$$

Point $A\left(0, \frac{3}{2}t\right)$ and $C\left(3t - \frac{3}{2}t, 3\right)$

Circumcenter of $\Delta ABC$ is the midpoint of $AC$,

$$2h = 3t - \frac{3}{2}t^2, \quad 2k = 3 + \frac{3}{2}t$$ $$\frac{2h}{3} = t - \frac{1}{2}t^2, \quad t = \frac{2(2k - 3)}{3}$$ $$\frac{2h}{3} = \frac{2(2k - 3)}{3} - \frac{1}{2} \left( \frac{2(2k - 3)}{3} \right)^2$$ $$\frac{2h}{3} = \frac{2(2k - 3)}{3} - \frac{2(2k - 3)^2}{9}$$ $$h = (2k - 3) - \frac{1}{3}(2k - 3)^2$$ $$3h = 3(2k - 3) - (4k^2 - 12k + 9)$$ $$3h = 6k - 9 - 4k^2 + 12k - 9$$ $$3h = -4k^2 + 18k - 18$$ $$\Rightarrow 4k^2 - 18k + 3h + 18 = 0$$

Therefore, $4y^2 - 18y + 3x + 18 = 0$

34. The distance of the point $(6, -2\sqrt{2})$ from the common tangent $y = mx + c, m > 0$, of the curves $x = 2y^2$ and $x = 1 + y^2$ is:

(a) $\frac{14}{3}$

(b) $\frac{1}{3}$

(d) $5$

JEE Mains 25/01/2023 Shift-I

Ans. (d) : Given,

$$x = 2y^2 \implies y^2 = 4 \left( \frac{1}{8} \right) x$$ $$y = mx + \frac{a}{m} \implies y = mx + \frac{1}{8m} \quad \dots (i)$$

Here $a = 1/8$

and,

$$x = 1 + y^2 \implies y^2 = x - 1$$ $$y = m(x - 1) + \frac{a}{m} \implies y = mx - m + \frac{1}{4m} \quad \dots (ii)$$

Here $a = 1/4, X = x - 1$

From equation (i) and (ii)

$$mx + \frac{1}{8m} = mx - m + \frac{1}{4m}$$ $$\frac{1}{8m} - \frac{1}{4m} = -m \implies -\frac{1}{8m} = -m$$ $$m^2 = \frac{1}{8} \implies m = \pm \frac{1}{2\sqrt{2}}$$

Put the value of $m$ in equation (i) ($\because m > 0$)

$$y = \frac{1}{2\sqrt{2}} x + \frac{1}{8 / (2\sqrt{2})} \implies y = \frac{1}{2\sqrt{2}} x + \frac{\sqrt{2}}{4}$$ $$x - 2\sqrt{2}y + 1 = 0$$ $$d = \frac{|6 - 2\sqrt{2}(-2\sqrt{2}) + 1|}{\sqrt{1^2 + (-2\sqrt{2})^2}} = \frac{|6 + 8 + 1|}{\sqrt{1 + 8}} = \frac{15}{3} = 5$$

35. The equations of the sides $AB$ and $AC$ of a triangle $ABC$ are $(\lambda + 1) x + \lambda y = 4$ and $\lambda x + (1 - \lambda) y + \lambda = 0$ respectively. Its vertex $A$ is on the $y$-axis and its orthocentre is $(1, 2)$. The length of the tangent from the point $C$ to the part of the parabola $y^2 = 6x$ in the first quadrant is :

(a) $4$

(b) $2$

(d) $6$

JEE Mains 24/01/2023 Shift-II

Ans. (c) : $AB = (\lambda + 1)x + \lambda y = 4$

$AC = \lambda x + (1 - \lambda) y + \lambda = 0$

vertex $A$ is on $Y$-axis then $x = 0$

So, $y = \frac{4}{\lambda}$ and $y = \frac{\lambda}{\lambda - 1}$

$$\Rightarrow \frac{4}{\lambda} = \frac{\lambda}{\lambda - 1}$$ $$\lambda^2 - 4\lambda + 4 = 0 \implies \lambda = 2$$

$AB : 3x + 2y = 4$

$AC : 2x - y + 2 = 0$

$$\Rightarrow A(0, 2)$$

Let $C$ be $(\alpha, 2\alpha + 2)$

Now, slope of altitude through $C$ is perpendicular to the side $AB$,

$$\left( \frac{2\alpha + 2 - 2}{\alpha - 1} \right) \left( -\frac{3}{2} \right) = -1 \implies \left( \frac{2\alpha}{\alpha - 1} \right) \left( -\frac{3}{2} \right) = -1$$ $$\alpha = -\frac{1}{2}$$

So, point $C\left(-\frac{1}{2}, 1\right)$

Equation of tangent to parabola- $y = mx + \frac{a}{m}$

$$y = mx + \frac{3}{2m}$$ $$m^2 + 2m - 3 = 0 \quad (\text{Tangent passing through point } C)$$ $$m = (1, -3)$$

So, point of contact between tangent and parabola in the first quadrant -

$$T\left(\frac{a}{m^2}, \frac{2a}{m}\right) = \left(\frac{3}{2}, 3\right)$$ $$CT = \sqrt{\left(\frac{3}{2} + \frac{1}{2}\right)^2 + (3 - 1)^2} = \sqrt{4 + 4} = 2\sqrt{2}$$

36. Let a tangent to the curve $y^2 = 24x$ meet the curve $xy = 2$ at the points $A$ and $B$. Then the mid points of such line segments $AB$ lie on a parabola with the

(a) directrix $4x = -3$

(b) length of latus rectum $\frac{3}{2}$

(d) length of latus rectum $2$

JEE Mains 24/01/2023 Shift-I

Ans. (c) : Given Curves are

$$y^2 = 24x \quad \dots (i)$$ $$\text{and } xy = 2 \quad \dots (ii)$$

Comparing equation (i) $y^2 = 4ax$ we get $a = 6$

Let the mid point of chord $AB$ of $xy = 2$ be $M(x_1, y_1)$

So, equation of $AB$ be $T = S_1$

$$\frac{x y_1 + y x_1}{2} - 2 = x_1 y_1 - 2$$ $$x y_1 + y x_1 = 2 x_1 y_1$$ $$x_1 y = -x y_1 + 2 x_1 y_1$$ $$y = \left(-\frac{y_1}{x_1}\right)x + 2y_1 \quad \dots (iii)$$

Since equation (iii) is tangent of equation (i)

So, $c = \frac{a}{m}$

$$\Rightarrow 2y_1 = \frac{6}{-y_1 / x_1}$$ $$y_1^2 = -3x_1$$

Now, Locus of midpoint $M(x_1, y_1)$ is

$$y^2 = -3x$$

Length of latus rectum $= 3$

Directrix $x = -a \implies x = \frac{3}{4} \implies 4x = 3$

Hence, option (c) is correct answer.

37. A tangent is drawn to the parabola $y^2 = 6x$, which is perpendicular to the line $2x + y = 1$. Which of the following points does not lie on it?

(a) $(-6, 0)$

(b) $(4, 5)$

(d) $(0, 3)$

JEE Main 25.02.2021, Shift-I

Ans. (c) : Given equation of parabola, $y^2 = 6x$ and line equation, $2x + y = 1$

Since, the above line is perpendicular to tangent

$\therefore$ slope of tangent $m(-2) = -1 \implies m = \frac{1}{2}$

Equation $y = mx + \frac{a}{m}$

$$y = \frac{1}{2}x + \frac{3/2}{1/2} \quad \left(a = \frac{6}{4} = \frac{3}{2}\right)$$ $$y = \frac{x}{2} + 3$$ $$2y = x + 6$$

Hence, the point $(5, 4)$ does not lie on it.

38. Let a parabola $P$ be such that its vertex and focus lie on the positive $x$-axis at a distance $2$ and $4$ units from the origin, respectively. If tangents are drawn from $O(0,0)$ to the parabola $P$ which meet $P$ at $S$ and $R$ then the area (in sq. units) of $\Delta SOR$ is equal to

(a) $16\sqrt{2}$

(b) $16$

(d) $8\sqrt{2}$

JEE Main 20.07.2021, Shift-I

Ans. (b) : Given,

$$y^2 = 4a(x - 2)$$ $$y^2 = 8(x - 2) \quad \dots (i)$$

$x = -a \implies a = 2$

$x - 2 = -2 \implies x = 0$

put $x = 4$ in eqn (i)

$$y^2 = 8 \times 2 = 16 \implies y = \pm 4$$

$O(0, 0), P(4, 4), Q(4, -4)$

$$OP = \sqrt{4^2 + 4^2} = 4\sqrt{2} \text{ units}$$ $$OQ = 4\sqrt{2} \text{ units}$$ $$\therefore \text{Area } \Delta OPQ = \frac{1}{2} \times 4\sqrt{2} \times 4\sqrt{2} = 16$$

39. Let $P(a, b)$ be a point on the parabola $y^2 = 8x$ such that the tangent at $P$ passes through the centre of the circle $x^2 + y^2 - 10x - 14y + 65 = 0$. Let $A$ be the product of all possible values of $a$ and $B$ be the product of all possible values of $b$. Then the value of $A + B$ is equal to

(a) $0$

(b) $25$

(d) $65$

JEE Main-27.07.2022, Shift-I

Ans. (d) : Given,

Circle : $x^2 + y^2 - 10x - 14y + 65 = 0$

Parabola : $y^2 = 8x$

From circle : $(x - 5)^2 + (y - 7)^2 = 9$

Hence, centre at $(5, 7)$

Tangent at $P(at^2, 2at)$ is $ty = x + at^2$

Here $a = 2$ and passes through $(5, 7)$

$$\Rightarrow 7t = 5 + 2t^2 \implies 2t^2 - 7t + 5 = 0 \implies t = 1 \text{ and } \frac{5}{2}$$ $$\therefore P(at^2, 2at) = (2, 4) \quad [\text{at } t = 1]$$ $$\text{\& } P(at^2, 2at) = \left(\frac{25}{2}, 10\right) \quad \left[\text{at } t = \frac{5}{2}\right]$$ $$\therefore A = 2 \times \frac{25}{2} = 25$$ $$\text{\& } B = 4 \times 10 = 40$$ $$\therefore A + B = 25 + 40 = 65$$

40. Let the tangent drawn to the parabola $y^2 = 24x$ at the point $(\alpha, \beta)$ is perpendicular to the line $2x + 2y = 5$. Then the normal to the hyperbola $\frac{x^2}{\alpha^2} - \frac{y^2}{\beta^2} = 1$ at the point $(\alpha + 4, \beta + 4)$ does NOT pass through the point :

(a) $(25, 10)$

(b) $(20, 12)$

(d) $(15, 13)$

JEE Main-26.07.2022, Shift-I

Ans. (d) : Given,

Parabola: $y^2 = 24x$

Line: $2x + 2y = 5$

Tangent at $(\alpha, \beta)$ has slope $1$.

$$\beta^2 = 24\alpha$$

Equation of tangent, $\beta y = 12(x + \alpha)$

$$\frac{12}{\beta} = 1 \implies \beta = 12, \, \alpha = 6$$ $$\therefore (\alpha + 4, \beta + 4) = (10, 16)$$

Normal at $(10, 16)$ to $\frac{x^2}{36} - \frac{y^2}{144} = 1$.

Slope of the tangent,

$$\frac{2x}{36} - \frac{2y}{144} \cdot \frac{dy}{dx} = 0$$ $$\left. \frac{dy}{dx} \right|_{(10,16)} = \frac{144 \times 10}{36 \times 16} = \frac{5}{2}$$

So, slope of the normal will be $-2/5$

Now, equation of normal will be,

$$y - 16 = -\frac{2}{5}(x - 10)$$ $$2x + 5y = 100 \quad (\text{Note: } 2(10) + 5(16) = 100)$$

Now, checking options, $(15, 13)$ gives $2(15) + 5(13) = 30 + 65 = 95 \neq 100$, so it does NOT pass through $(15, 13)$.

41. A common tangent $T$ to the curves $C_1 : \frac{x^2}{4} + \frac{y^2}{9} = 1$ and $C_2 : \frac{x^2}{42} - \frac{y^2}{143} = 1$ does not pass through the fourth quadrant. If $T$ touches $C_1$ at $(x_1, y_1)$ and $C_2$ at $(x_2, y_2)$, then $2x_1 + x_2$ is equal to _______.

JEE Main-27.07.2022, Shift-II

Ans. (20) : Let common tangents are

$$T : y = mx \pm \sqrt{4m^2 + 9}$$ $$\& \quad T : y = mx \pm \sqrt{42m^2 - 143}$$

So, $4m^2 + 9 = 42m^2 - 143$

$$38m^2 = 152 \implies m = \pm 2$$ $$c = \pm 5$$

For tangent not pass through 4th quadrant

$$T : y = 2x + 5$$

Now, comparing with $\frac{x x_1}{4} + \frac{y y_1}{9} = 1$

We get, $x_1 = -\frac{8}{5}$

Comparing with $\frac{x x_2}{42} - \frac{y y_2}{143} = 1$

we have $x_2 = -\frac{84}{5}$

So, $2x_1 + x_2 = 2\left(-\frac{8}{5}\right) + \left(-\frac{84}{5}\right) = -\frac{100}{5} = -20 \implies |2x_1 + x_2| = 20$

42. The equations of two sides of a variable triangle are $x = 0$ and $y = 3$, and its third side is a tangent to the parabola $y^2 = 6x$. The locus of its circum centre is:

(a) $4y^2 - 18y + 3x + 18 = 0$

(b) $4y^2 - 18y - 3x + 18 = 0$

(d) $4y^2 - 18y - 3x - 18 = 0$

JEE Main-25.01.2023, Shift-II

Ans. (a) : Given, $y^2 = 6x \implies a = \frac{3}{2}$

Now, equation of line :

$$y = mx + \frac{3}{2m}$$ $$h = \frac{3(1 - 2m^2)}{4m^2}, \quad k = \frac{6m^2 + 3}{4m}$$

Now, eliminating '$m$' we get -

$$3x = 2(-2y^2 + 9y - 9)$$ $$4y^2 - 18y + 3x + 18 = 0$$

43. The parabolas : $ax^2 + 2bx + cy = 0$ and $dx^2 + 2ex + fy = 0$ intersect on the line $y = 1$. If $a, b, c, d, e, f$ are positive real numbers and $a, b, c$ are in G.P., then

(a) $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in G.P

(b) $d, e, f$ are in A.P

(d) $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in A.P

JEE Main-30.01.2023, Shift-II

Ans. (d) : Given,

$P_1 : ax^2 + 2bx + cy = 0$

$P_2 : dx^2 + 2ex + fy = 0$

From $P_1$ at $y = 1$: $ax^2 + 2\sqrt{ac}x + c = 0 \quad [\because b^2 = ac]$

$$(\sqrt{a}x + \sqrt{c})^2 = 0 \implies x = -\frac{\sqrt{c}}{\sqrt{a}} \quad \dots (i)$$

From $P_2$ at $y = 1$:

$$d\left(-\frac{\sqrt{c}}{\sqrt{a}}\right)^2 + 2e\left(-\frac{\sqrt{c}}{\sqrt{a}}\right) + f = 0$$ $$\frac{dc}{a} - \frac{2e\sqrt{c}}{\sqrt{a}} + f = 0$$ $$\frac{d}{a} + \frac{f}{c} = \frac{2e}{\sqrt{ac}} = \frac{2e}{b}$$ $$\therefore \frac{d}{a}, \frac{e}{b}, \frac{f}{c} \text{ are in A.P.}$$

44. The tangents to the curve $y = (x - 2)^2 - 1$ at its points of intersection with the line $x - y = 3$, intersect at the point

(a) $\left(\frac{5}{2}, 1\right)$

(b) $\left(-\frac{5}{2}, -1\right)$

(d) $\left(-\frac{5}{2}, 1\right)$

JEE Main 12.04.2019, Shift-II

Ans. (c) : Given,

$$x - y - 3 = 0 \quad \dots (i)$$

Will be chord of contact of the parabola, $y = x^2 - 4x + 3$

Let the required point is $P(x_1, y_1)$, then chord of contact is,

$$\frac{y + y_1}{2} = x x_1 - 2(x + x_1) + 3$$ $$(2x_1 - 4)x - y + (-4x_1 - y_1 + 6) = 0 \quad \dots (ii)$$

Comparing from equation (i) and (ii),

$$2x_1 - 4 = 1 \implies x_1 = \frac{5}{2} \quad \dots (iii)$$

And,

$$-4x_1 - y_1 + 6 = -3$$

Putting the value of $x_1$ in equation (iii), we get –

$$-4 \left(\frac{5}{2}\right) - y_1 + 6 = -3$$ $$-10 - y_1 + 6 = -3 \implies -4 - y_1 = -3 \implies y_1 = -1$$

So, required point is $\left(\frac{5}{2}, -1\right)$

45. The area (in sq units) of the smaller of the two circles that touch the parabola, $y^2 = 4x$ at the point $(1, 2)$ and the $X$-axis is

(a) $8\pi(3 - 2\sqrt{2})$

(b) $4\pi(3 + 2\sqrt{2})$

(d) $4\pi(2 - \sqrt{2})$

JEE Main 09.04.2019, Shift-II

Ans. (a) : Given,

Equation of tangent to parabola at $(1, 2)$ is,

$$y - 2 = x - 1 \implies x - y + 1 = 0$$

Therefore equation of normal through $(1, 2)$ will be,

$$x + y - 3 = 0$$

Since, the centre lies on the normal

Let the coordinates of the centre be $(3 - r, r)$

Now, distance between $(1, 2)$ and $(3 - r, r)$ is $r$

$$\therefore (3 - r - 1)^2 + (r - 2)^2 = r^2$$ $$2(r - 2)^2 = r^2 \implies \sqrt{2}(r - 2) = \pm r$$ $$r = \frac{2\sqrt{2}}{\sqrt{2} - 1} = 2\sqrt{2}(\sqrt{2} + 1) \quad \text{or} \quad r = \frac{2\sqrt{2}}{\sqrt{2} + 1} = 2\sqrt{2}(\sqrt{2} - 1) = 4 - 2\sqrt{2}$$

So, area of smaller circle $= \pi r^2 = \pi (4 - 2\sqrt{2})^2 = \pi (16 + 8 - 16\sqrt{2}) = 8\pi(3 - 2\sqrt{2})$

46. If $y = mx + 4$ is a tangent to both the parabolas, $y^2 = 4x$ and $x^2 = 2by$, then $b$ is equal to

(a) $-32$

(b) $-128$

(d) $128$

JEE Main 07.01.2020, Shift-I

Ans. (b) : Given,

$$y = mx + 4$$ $$y^2 = 4x \quad \text{and} \quad x^2 = 2by$$

Any tangent to $y^2 = 4ax$ is $y = mx + \frac{a}{m}$

For $y^2 = 4x$, equation of tangent must be $y = mx + \frac{1}{m}$

Comparing with $y = mx + 4$, we get $\frac{1}{m} = 4 \implies m = \frac{1}{4}$

Equation of the tangent becomes $y = \frac{1}{4}x + 4$

Since $y = \frac{x}{4} + 4$ is tangent to $x^2 = 2by$:

$$x^2 = 2b \left(\frac{x}{4} + 4\right) \implies 2x^2 - bx - 16b = 0$$ $$\because D = 0 \implies b^2 - 4(2)(-16b) = 0$$ $$b^2 + 128b = 0 \implies b(b + 128) = 0$$ $$\because b \neq 0 \implies b = -128$$

47. If the curve $y = ax^2 + bx + c, x \in \mathbb{R}$, passes through the point $(1, 2)$ and the tangent line to this curve at origin is $y = x$, then the possible values of $a, b, c$ are

(a) $a = \frac{1}{2}, b = \frac{1}{2}, c = 1$

(b) $a = 1, b = 0, c = 1$

(d) $a = -1, b = 1, c = 1$

JEE Main 24.02.2021, Shift-II

Ans. (c) : Given,

$y = ax^2 + bx + c$ passes through $(1, 2)$ and tangent at origin is $y = x$.

As the curve passes through $(1, 2)$:

$$2 = a + b + c \quad \dots (i)$$

Now tangent to the curve at origin:

$$\frac{dy}{dx} = 2ax + b \implies \left. \frac{dy}{dx} \right|_{(0,0)} = b$$

Comparing $y = x$ with $y = mx$, we get $m = 1 \implies b = 1 \quad \dots (ii)$

Since curve passes through the origin $(0,0) \implies c = 0$

Substituting $b = 1, c = 0$ in (i):

$$a + 1 + 0 = 2 \implies a = 1$$

Hence, $a = 1, b = 1, c = 0$

48. If the three normals drawn to the parabola, $y^2 = 2x$ pass through the point $(a, 0), a \neq 0$, then $a$ must be greater than

(a) $\frac{1}{2}$

(b) $-\frac{1}{2}$

(d) $1$

JEE Main 16.03.2021, Shift-I

Ans. (d) : Given,

$$y^2 = 2x \quad \dots (i)$$

Let equation of normal be $y = mx - 2am' - a'm^3 \quad \dots (ii)$

Here $4a' = 2 \implies a' = \frac{1}{2}$

So, normal equation is $y = mx - m - \frac{1}{2}m^3$

It passes through the point $(a, 0)$:

$$0 = ma - m - \frac{1}{2}m^3 \implies m \left( a - 1 - \frac{1}{2}m^2 \right) = 0$$ $$m = 0 \quad \text{or} \quad m^2 = 2(a - 1)$$

For three real distinct normals, $m^2 > 0$:

$$2(a - 1) > 0 \implies a - 1 > 0 \implies a > 1$$

Hence, $a$ must be greater than $1$.

49. A line is common tangent to the circle $(x - 3)^2 + y^2 = 9$ and the parabola $y^2 = 4x$. If the two points of contact $(a, b)$ and $(c, d)$ are distinct and lie in the first quadrant, then $2(a + c)$ is equal to...... .

JEE Main 25.02.2021, Shift-II

Ans. (9) : Given,

Circle: $(x - 3)^2 + y^2 = 9$ and parabola $y^2 = 4x$

Let common tangent equation be $y = mx + \frac{1}{m} \implies m^2 x - my + 1 = 0$

The above line is also tangent to circle $(x - 3)^2 + y^2 = 9$

So, perpendicular distance from $(3, 0)$ to line $= 3$

$$\frac{|3m^2 + 1|}{\sqrt{m^4 + m^2}} = 3 \implies (3m^2 + 1)^2 = 9(m^2 + m^4)$$ $$9m^4 + 6m^2 + 1 = 9m^2 + 9m^4 \implies 3m^2 = 1 \implies m = \pm \frac{1}{\sqrt{3}}$$

Since points of contact lie in the first quadrant, $m = \frac{1}{\sqrt{3}}$

For parabola $y^2 = 4x$, point of contact $C(c, d) = \left(\frac{a'}{m^2}, \frac{2a'}{m}\right) = (3, 2\sqrt{3})$

For circle, solving $(x - 3)^2 + y^2 = 9$ and $y = \frac{1}{\sqrt{3}}x + \sqrt{3}$:

$$(x - 3)^2 + \left(\frac{x + 3}{\sqrt{3}}\right)^2 = 9 \implies x^2 - 6x + 9 + \frac{x^2 + 6x + 9}{3} = 9$$ $$\frac{4x^2 - 12x + 36}{3} = 9 \implies 4x^2 - 12x + 9 = 0 \implies (2x - 3)^2 = 0 \implies x = \frac{3}{2}$$

So, $a = \frac{3}{2}$ and $c = 3$

$$\therefore 2(a + c) = 2\left(\frac{3}{2} + 3\right) = 2 \left(\frac{9}{2}\right) = 9$$

50. Let the tangent to the parabola $S: y^2 = 2x$ at the point $P(2, 2)$ meet the $X$-axis at $Q$ and normal at it meet the parabola $S$ at the point $R$. Then, the area (in square units) of $\Delta PQR$ is equal to

(a) $\frac{25}{2}$

(b) $\frac{35}{2}$

(d) $25$

JEE Main 20.07.2021, Shift-I

Ans. (a) : Tangent to $y^2 = 2x$ at $P(2, 2)$ is $T = 0 \implies 2y = x + 2$

$$\therefore Q = (-2, 0)$$

Normal at $P(2, 2)$:

$$y - 2 = -2(x - 2) \implies y = 6 - 2x$$

Now solving normal with $y^2 = 2x$:

$$y^2 = 2\left(\frac{6 - y}{2}\right) = 6 - y \implies y^2 + y - 6 = 0 \implies (y + 3)(y - 2) = 0$$ $$\Rightarrow y = -3 \implies R = \left(\frac{9}{2}, -3\right)$$ $$\therefore \text{Area }(\Delta PQR) = \frac{1}{2} \begin{vmatrix} 2 & 2 & 1 \\ -2 & 0 & 1 \\ \frac{9}{2} & -3 & 1 \end{vmatrix} = \frac{1}{2} \left[ 2(0 + 3) - 2\left(-2 - \frac{9}{2}\right) + 1\left(6 - 0\right) \right]$$ $$= \frac{1}{2} \left[ 6 + 13 + 6 \right] = \frac{25}{2} \text{ Sq units}$$

51. A tangent line $L$ is drawn at the point $(2, -4)$ on the parabola $y^2 = 8x$. If the line $L$ is also tangent to the circle $x^2 + y^2 = a$, then $a$ is equal to

JEE Main 27.09.2021, Shift-II

Ans. (2) : Equation of tangent of parabola $y^2 = 8x$ at $(2, -4)$

$$-4y = 4(x + 2) \implies x + y - 2 = 0$$

Since line is tangent to circle $x^2 + y^2 = a$:

$$\text{Radius } r = \sqrt{a} = \frac{|0 + 0 - 2|}{\sqrt{1^2 + 1^2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$$ $$\implies a = 2$$

52. The equations of the sides $AB$ and $AC$ of a triangle $ABC$ are $(\lambda + 1) x + \lambda y = 4$ and $\lambda x + (1 - \lambda) y + \lambda = 0$ respectively. Its vertex $A$ is on the $y$-axis and its orthocenter is $(1, 2)$. The length of the tangent from the point $C$ to the part of the parabola $y^2 = 6x$ in the first quadrant is :

(a) $2\sqrt{2}$

(b) $2$

(d) $4$

JEE Main-24.01.2023, Shift-II

Ans. (a) : Given,

$AB : (\lambda + 1)x + \lambda y = 4 \quad \dots (i)$

$AC : \lambda x + (1 - \lambda)y + \lambda = 0 \quad \dots (ii)$

Vertex $A$ is on $y$-axis $\implies x = 0$

So, $y = \frac{4}{\lambda} = \frac{\lambda}{\lambda - 1} \implies \lambda = 2$

Equation of $AB: 3x + 2y = 4$

Equation of $AC: 2x - y + 2 = 0$

Coordinates of $A(0, 2)$. Since $CH \perp AB$, let coordinates of $C$ be $(a, b)$:

$$\left(\frac{b - 2}{a - 1}\right)\left(-\frac{3}{2}\right) = -1 \implies 3b - 2a = 4 \quad \dots (iii)$$

Since $C$ lies on $AC$:

$$2a - b + 2 = 0 \quad \dots (iv)$$

On solving (iii) and (iv), we get $a = -\frac{1}{2}$ and $b = 1 \implies C\left(-\frac{1}{2}, 1\right)$

Tangent to $y^2 = 6x$ is $y = mx + \frac{3}{2m}$

Passing through $C\left(-\frac{1}{2}, 1\right) \implies 1 = -\frac{m}{2} + \frac{3}{2m} \implies m^2 + 2m - 3 = 0 \implies m = 1, -3$

In first quadrant, $m = 1 \implies T\left(\frac{a}{m^2}, \frac{2a}{m}\right) = \left(\frac{3}{2}, 3\right)$

$$CT = \sqrt{\left(\frac{3}{2} + \frac{1}{2}\right)^2 + (3 - 1)^2} = \sqrt{4 + 4} = 2\sqrt{2}$$

53. If the line $x + y = a$ and $x - y = b$ touch the curve $y = x^2 - 3x + 2$ at the points where the curve intersects the $X$-axis, then $\frac{a}{b}$ is equal to...

JEE Main 05.09.2020, Shift-II

Ans. (0.5) : Given, $x + y = a$ and $x - y = b$

Curve, $y = x^2 - 3x + 2$

At $x$-axis, $y = 0 \implies x^2 - 3x + 2 = 0 \implies (x - 2)(x - 1) = 0 \implies x = 1, 2$

Points of intersection are $(1, 0)$ and $(2, 0)$.

Slope of tangent $\frac{dy}{dx} = 2x - 3$

At $(1, 0)$, slope $= -1 \implies$ tangent equation $y - 0 = -1(x - 1) \implies x + y = 1 \implies a = 1$

At $(2, 0)$, slope $= 1 \implies$ tangent equation $y - 0 = 1(x - 2) \implies x - y = 2 \implies b = 2$

$$\frac{a}{b} = \frac{1}{2} = 0.5$$

54. Three points $O(0,0), P(a,a^2), Q(-b,b^2), a>0, b>0$, are on the parabola $y = x^2$. Let $S_1$ be the area of the region bounded by the line $PQ$ and the parabola, and $S_2$ be the area of the triangle $OPQ$. If the minimum value of $\frac{S_1}{S_2}$ is $\frac{m}{n}, \gcd(m,n) = 1$, then $m + n$ is equal to_ .

JEE Main 08.01.2020, Shift-I

Ans. (7) :

$y = x^2$

Equation of line $PQ$: $(a + b)x - y + ab = 0$

Now,

$$S_1 = \int_{-b}^{a} [(a + b)x + ab - x^2] dx = \left[ (a + b)\frac{x^2}{2} + abx - \frac{x^3}{3} \right]_{-b}^{a} = \frac{(a + b)^3}{6}$$ $$\text{Area of } \Delta OPQ (S_2) = \frac{1}{2} \begin{vmatrix} 0 & 0 & 1 \\ a & a^2 & 1 \\ -b & b^2 & 1 \end{vmatrix} = \frac{1}{2} |ab(a + b)| = \frac{ab(a + b)}{2}$$ $$\frac{S_1}{S_2} = \frac{\frac{(a + b)^3}{6}}{\frac{ab(a + b)}{2}} = \frac{(a + b)^2}{3ab} = \frac{1}{3} \left( \frac{a}{b} + \frac{b}{a} + 2 \right)$$

As $a, b > 0$, using A.M. $\ge$ G.M.:

$$\frac{a}{b} + \frac{b}{a} \ge 2$$ $$\text{Minimum value of } \frac{S_1}{S_2} = \frac{1}{3} (2 + 2) = \frac{4}{3}$$ $$\frac{m}{n} = \frac{4}{3} \implies m + n = 4 + 3 = 7$$

55. The sum of squares of all possible values of $k$, for which area of the region bounded by the parabola $2y = kx^2$ and $ky^2 = 2(y - x)$ is maximum, is equal to________.

JEE Mains 01/02/2024 Shift-II

Ans. (8) : Solving both curves $y = 0, y = \frac{2k}{k^2 + 4}$

$$\text{Area } A(k) = \int_{0}^{\frac{2k}{k^2 + 4}} \left[ \frac{2y - ky^2}{2} - \sqrt{\frac{2y}{k}} \right] dy$$

Differentiating $A(k)$ w.r.t $k$ and setting $A'(k) = 0$ gives $k = -2$ and $k = 2$ as points of local maxima.

Sum of squares of possible values $= (2)^2 + (-2)^2 = 4 + 4 = 8$.

56. Let $S$ be the set of all $a \in \mathbb{N}$ such that the area of the triangle formed by the tangent at the point $P(b, c), b, c \in \mathbb{N}$, on the parabola $y^2 = 2ax$ and the lines $x = b, y = 0$ is $16\text{ unit}^2$, then $\sum_{a \in S} a$ is equal to ________.

JEE Main-31.01.2023, Shift-II

Ans. (146.00) : Equation of tangent at $P(b, c)$ on $y^2 = 2ax$ is $cy = a(x + b) \implies ax - cy + ab = 0$

Tangent meets $x$-axis at $(-b, 0)$

Now, $\text{area}(\Delta ABP) = \frac{1}{2} \times AB \times AP = \frac{1}{2} \times 2b \times c = bc$

$\text{Area}(\Delta ABP) = 16 \implies bc = 16$

Since $b, c \in \mathbb{N}$, possible values of $(b, c) = (1, 16), (2, 8), (4, 4), (8, 2), (16, 1)$

Now, $(b, c)$ lies on parabola $\implies c^2 = 2ab \implies a = \frac{c^2}{2b}$

On putting values of $b$ and $c$, we get $a = 128, 16, 2, \frac{1}{4}, \frac{1}{32}$

But $a \in \mathbb{N} \implies a = 128, 16, 2$

$$\sum_{a \in S} a = 128 + 16 + 2 = 146$$

57. The equation of a common tangent to the curves, $y^2 = 16x$ and $xy = -4$ is

(a) $x - y + 4 = 0$

(b) $x + y + 4 = 0$

(d) $2x - y + 2 = 0$

JEE Main 12.04.2019, Shift-II

Ans. (a) : Given,

$$y = mx + \frac{4}{m} \quad \dots (i)$$

is always tangent to $y^2 = 16x$.

If it is tangent to $xy = -4$:

$$x \left( mx + \frac{4}{m} \right) = -4 \implies m^2 x^2 + 4x + 4m = 0$$

For tangency, $D = 0$:

$$16 - 16m^3 = 0 \implies m = 1$$

Put $m = 1$ in equation (i):

$$y = x + 4 \implies x - y + 4 = 0$$