JEE Advanced 2024 Paper 2 Q07 Mathematics Coordinate Geometry Parabola Medium

JEE Advanced 2024 Paper 2 · Q07 · Parabola

Let $A_{1}, B_{1}, C_{1}$ be three points in the $xy$-plane. Suppose that the lines $A_{1}C_{1}$ and $B_{1}C_{1}$ are tangents to the curve $y^{2} = 8x$ at $A_{1}$ and $B_{1}$, respectively. If $O = (0, 0)$ and $C_{1} = (-4, 0)$, then which of the following statements is (are) TRUE?

  1. A. The length of the line segment $OA_{1}$ is $4\sqrt{3}$
  2. B. The length of the line segment $A_{1}B_{1}$ is $16$
  3. C. The orthocenter of the triangle $A_{1}B_{1}C_{1}$ is $(0, 0)$
  4. D. The orthocenter of the triangle $A_{1}B_{1}C_{1}$ is $(1, 0)$
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, C

Solution

Parametrise points on $y^{2} = 8x$ as $(2t^{2}, 4t)$. Tangent at $(2t^{2}, 4t)$ is $ty = x + 2t^{2}$. Passes through $C_{1} = (-4, 0)$: $0 = -4 + 2t^{2} \Rightarrow t^{2} = 2 \Rightarrow t = \pm\sqrt{2}$. So $A_{1} = (4, 4\sqrt{2})$ and $B_{1} = (4, -4\sqrt{2})$. Then $|OA_{1}| = \sqrt{16 + 32} = \sqrt{48} = 4\sqrt{3}$ — (A) TRUE. $|A_{1}B_{1}| = 8\sqrt{2} \ne 16$ — (B) FALSE. For the orthocentre: $A_{1}B_{1}$ is the vertical line $x = 4$; altitude from $C_{1}$ is horizontal $y = 0$. Slope of $A_{1}C_{1} = (4\sqrt{2})/(8) = \sqrt{2}/2$; altitude from $B_{1}$ has slope $-\sqrt{2}$ and passes through $(4, -4\sqrt{2})$: $y + 4\sqrt{2} = -\sqrt{2}(x-4)$. At $y = 0$: $4\sqrt{2} = -\sqrt{2}(x-4) \Rightarrow x = 0$. Orthocentre $= (0, 0)$ — (C) TRUE; (D) FALSE.

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