Let A_1, B_1, C_1 be three points in the x y -plane. Suppose that the lines A_1 C_1 and B_1 C_1 are tangents to the curve y^2=8 x at A_1 and B_1 , respectively. If O=(0,0) and C_1=(-4,0) , then which of the following

Correct answer: A. The length of the line segment O A_1 is 4 3 Equation of tangent at (2 t^2, 4 t) is ty=x+2 t^2 It is passing through (-4,0) 0=-4+2 t^2 t= ± 2 . array rl A_1=(4,4 2 ) B_1=(4,-4 2 ) array \ aligned & OA_1= 48 =4 3 & ~A_1 ~B_1=8 2 aligned Equation of altitude of Δ A_1 B_1 C_1 drawn from A_1 is y-4 2 = 2 (x-4) 2 x-y=0 ...(1) Equation of altitude of Δ A_1 B_1 C_1 drawn from C_1 is x=0 ...(2) Solving (1) and (2) orthocentre is (0,0) correct options are (1), (3)

JEE Advanced 2024Paper 2

Exam: JEE Advanced
Year: 2024
Paper: Paper 2
Subject: Mathematics
Chapter: Parabola

JEE Advanced 2024 · Paper 2 · Mathematics · Parabola

Let A_1, B_1, C_1 be three points in the x y -plane. Suppose that...

Let

A_1, B_1, C_1

be three points in the

x y

-plane. Suppose that the lines

A_1 C_1

and

B_1 C_1

are tangents to the curve

y^2=8 x

A_1

and

B_1

, respectively. If

O=(0,0)

and

C_1=(-4,0)

, then which of the following statements is (are) TRUE? Let A_1, B_1, C_1 be three points in the x y -plane. Suppose that the lines A_1 C_1 and B_1 C_1 are tangents to the

Let A_1, B_1, C_1 be three points in the x y -plane. Suppose that the lines A_1 C_1 and B_1 C_1 are tangents to the

Options

AThe length of the line segment $O A_1$ is $4 \sqrt{3}$
BThe length of the line segment $A_1 B_1$ is 16
CThe orthocenter of the triangle $A_1 B_1 C_1$ is $(0,0)$
DThe orthocenter of the triangle $A_1 B_1 C_1$ is $(1,0)$

Correct answer

A. The length of the line segment $O A_1$ is $4 \sqrt{3}$

Step-by-step solution

Equation of tangent at

\left(2 t^2, 4 t\right)

\mathrm{ty}=\mathrm{x}+2 \mathrm{t}^2 \because

It is passing through

(-4,0) \therefore 0=-4+2 \mathrm{t}^2 \Rightarrow \mathrm{t}= \pm \sqrt{2} \left.

\begin{array}{rl}\therefore \mathrm{A}_1=(4,4 \sqrt{2}) \\ \mathrm{B}_1=(4,-4 \sqrt{2})\end{array}

\right\}

\begin{aligned} & \mathrm{OA}_1=\sqrt{48}=4 \sqrt{3} \\ & \mathrm{~A}_1 \mathrm{~B}_1=8 \sqrt{2}\end{aligned}

Equation of altitude of

\Delta A_1 B_1 C_1

drawn from

A_1

is

y-4 \sqrt{2}=\sqrt{2}(x-4) \Rightarrow \sqrt{2} x-y=0

...(1) Equation of altitude of

\Delta A_1 B_1 C_1

drawn from

C_1

is

\mathrm{x}=0

...(2) Solving (1) and (2)

\Rightarrow

orthocentre is

(0,0) \therefore$ correct options are (1), (3)

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