JEE Advanced 2021 Paper 2 Q06 Mathematics Coordinate Geometry Parabola Medium

JEE Advanced 2021 Paper 2 · Q06 · Parabola

Let $E$ denote the parabola $y^2 = 8x$. Let $P = (-2, 4)$, and let $Q$ and $Q'$ be two distinct points on $E$ such that the lines $PQ$ and $PQ'$ are tangents to $E$. Let $F$ be the focus of $E$. Then which of the following statements is (are) TRUE?

  1. A. The triangle $PFQ$ is a right-angled triangle
  2. B. The triangle $QPQ'$ is a right-angled triangle
  3. C. The distance between $P$ and $F$ is $5\sqrt{2}$
  4. D. $F$ lies on the line joining $Q$ and $Q'$
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, B, D

Solution

For $y^2 = 8x$, $a = 2$, $F = (2,0)$. Take $Q = (2t_1^2, 4t_1)$, $Q' = (2t_2^2, 4t_2)$. Since $P=(-2,4)$ lies on the directrix $x=-2$, $P$ lies on the directrix, so $t_1 t_2 = -1$ (chord-of-contact passes through focus) and $t_1+t_2 = 2$ (from $P$ being on both tangents). (A) Slope$(PF)\cdot$slope$(FQ) = -1$ (since the chord of contact from a point on the directrix subtends a right angle at the focus). TRUE. (B) Slope$(PQ)\cdot$slope$(PQ') = -\dfrac{1}{t_1}\cdot -\dfrac{1}{t_2} = \dfrac{1}{t_1 t_2} = -1$. TRUE. (C) $PF = \sqrt{16+16} = 4\sqrt{2}\ne 5\sqrt{2}$. FALSE. (D) Chord of contact from $P=(-2,4)$: $4y = 4(x-2)$, i.e. $y = x-2$, passes through $F=(2,0)$. TRUE.

Want to drill deeper? Open this question inside SolveKar AI and tap "Why?" on any step — the AI Mentor reads your full solution context and explains until it clicks. Download SolveKar AI →