Mathematics · Parabola

JEE Advanced 2024 — Paper 2 — Question 7

Let A1,B1,C1A_{1}, B_{1}, C_{1} be three points in the xyx y-plane. Suppose that the lines A1C1A_{1} C_{1} and B1C1B_{1} C_{1} are tangents to the curve y2=8xy^{2}=8 x at A1A_{1} and B1B_{1}, respectively. If O=(0,0)O=(0,0) and C1=(−4,0)C_{1}=(-4,0), then which of the following statements is(are) TRUE?

  1. Option A:

    The length of the line segment OA1\mathrm{OA}_{1} is 434 \sqrt{3}

    Correct
  2. Option B:

    The length of the line segment A1B1A_{1} B_{1} is 16

  3. Option C:

    The orthocenter of the triangle A1 B1C1\mathrm{A}_{1} \mathrm{~B}_{1} \mathrm{C}_{1} is (0,0)(0,0)

    Correct
  4. Option D:

    The orthocenter of the triangle A1B1C1A_{1} B_{1} C_{1} is (1,0)(1,0)

Answer: A, C

Step-by-step solution

Let OA1\mathrm{OA}_{1} be y=mx+c\mathrm{y}=\mathrm{mx}+\mathrm{c}

⇒0=−4 m+c\Rightarrow 0=-4 \mathrm{~m}+\mathrm{c} or c=4 m\mathrm{c}=4 \mathrm{~m}

Also, 2=mc2=\mathrm{mc}

∴m=±12,c=±22\therefore \mathrm{m}= \pm \frac{1}{\sqrt{2}}, \mathrm{c}= \pm 2 \sqrt{2}

1t=m\frac{1}{\mathrm{t}}=\mathrm{m}

⇒A1(4,42)\Rightarrow \mathrm{A}_{1}(4,4 \sqrt{2}) and B1(4,−42)\mathrm{B}_{1}(4,-4 \sqrt{2})

Orthocentre is (0,0)(0,0)

A1 B1=82\mathrm{A}_{1} \mathrm{~B}_{1}=8 \sqrt{2}

OA1=43\mathrm{OA}_{1}=4 \sqrt{3}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2024
Paper
Paper 2
Subject
Mathematics
Chapter
Parabola
Topic
Pole and Polar wrt a Parabola
Let A 1 , B 1 , C 1 be three points in the x y -plane. Suppose that… | JEE Advanced 2024 PYQ with Solution · DhiX AI