JEE Advanced 2023 Paper 1 Q02 Mathematics Coordinate Geometry Parabola Hard

JEE Advanced 2023 Paper 1 · Q02 · Parabola

Let $T_1$ and $T_2$ be two distinct common tangents to the ellipse $E: \dfrac{x^2}{6} + \dfrac{y^2}{3} = 1$ and the parabola $P: y^2 = 12x$. 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?

  1. A. The area of the quadrilateral $A_1 A_2 A_3 A_4$ is $35$ square units
  2. B. The area of the quadrilateral $A_1 A_2 A_3 A_4$ is $36$ square units
  3. C. The tangents $T_1$ and $T_2$ meet the $x$-axis at the point $(-3, 0)$
  4. D. The tangents $T_1$ and $T_2$ meet the $x$-axis at the point $(-6, 0)$
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, C

Solution

For $y^2 = 12x$, tangent in slope form: $y = mx + 3/m$. Imposing common-tangent condition with the ellipse $\dfrac{x^2}{6}+\dfrac{y^2}{3}=1$: $9/m^2 = 6m^2 + 3 \Rightarrow 2m^4 + m^2 - 3 = 0 \Rightarrow m^2 = 1 \Rightarrow m = \pm 1$. Common tangents are $y = x + 3$ and $y = -x - 3$, both passing through $(-3, 0)$ on the $x$-axis. The contact points: $A_1 = (3, 6)$, $A_4 = (3, -6)$ on $P$; $A_2 = (-2, 1)$, $A_3 = (-2, -1)$ on $E$. The quadrilateral is a trapezium with parallel sides $A_1 A_4 = 12$ and $A_2 A_3 = 2$, height $= 5$. Area $= \dfrac{1}{2}(12+2)(5) = 35$.

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