JEE Advanced 2023 — Paper 1 — MATH-I solution

JEE Advanced2023Paper 1MATH-I

Let $T_{1}$ and $T_{2}$ be two distinct common tangents to the ellipse $E : \frac{x ^{2}}{6} + \frac{y ^{2}}{3} = 1$ and the parabola $P : y^{2} = 12 x$ . Suppose that the tangent $T_{1}$ touches $P$ and $E$ at the points $A_{1}$ and $A_{2}$ , respectively and the tangent $T_{2}$ touches $P$ and $E$ at the points $A_{4}$ and $A_{3}$ , respectively. Then which of the following statements is(are) true?

A
The area of the quadrilateral $A_{1} A_{2} A_{3} A_{4}$ is $35$ square units
B
The area of the quadrilateral $A_{1} A_{2} A_{3} A_{4}$ is $36$ square units
C
The tangents $T_{1}$ and $T_{2}$ meet the $x$ -axis at the point $(- 3, 0)$
D
The tangents $T_{1}$ and $T_{2}$ meet the $x$ -axis at the point $(- 6, 0)$

Solution

For the parabola $y^{2} = 12 x$ (so $4 a = 12$ , $a = 3$ ), a tangent of slope $m$ has the form $y = m x + \frac{3}{m}$ .

This line is tangent to the ellipse $\frac{x ^{2}}{6} + \frac{y ^{2}}{3} = 1$ iff $c^{2} = a_{e}^{2} m^{2} + b_{e}^{2}$ , with $a_{e}^{2} = 6$ , $b_{e}^{2} = 3$ and $c = \frac{3}{m}$ :

$\frac{9}{m ^{2}} = 6 m^{2} + 3 \Rightarrow 6 m^{4} + 3 m^{2} - 9 = 0 \Rightarrow m^{2} = 1$ .

So the two common tangents are $T_{1} : y = x + 3$ and $T_{2} : y = - x - 3$ , both of which cut the $x$ -axis at $(- 3, 0)$ .

Point of contact with $P$ for slope $m$ is $(\frac{a}{m ^{2}}, \frac{2 a}{m})$ , giving $A_{1} (3, 6)$ and $A_{4} (3, - 6)$ .

Point of contact with $E$ for slope $m$ on $\frac{x ^{2}}{6} + \frac{y ^{2}}{3} = 1$ is $(- \frac{a _{e}^{2} m}{c}, \frac{b _{e}^{2}}{c})$ , giving $A_{2} (- 2, 1)$ and $A_{3} (- 2, - 1)$ .

The quadrilateral $A_{1} A_{2} A_{3} A_{4}$ is a trapezium with parallel vertical sides $A_{1} A_{4} = 12$ and $A_{2} A_{3} = 2$ , separated by horizontal distance $5$ .

Area $= \frac{1}{2} (12 + 2) (5) = 35$ sq. units.

$∴$ The correct options are (A) and (C).

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