A normal with slope 1 6 is drawn from the point (0,-α) to the parabola x^2=-4 a y , where a>0 . Let L be the line passing through (0,-α) and parallel to the directrix of the parabola. Suppose that L intersects the

Correct answer: 0 (d y)/(d x)=.(x)/(-2 a) (d y)/(d x)|_N=-t Slope of normal =(1)/(t)= 1 6 t= 6 Now, -at^2+α 2 at = 1 t -at^2+α=2 a -6 a+α=2 a α=8 a For A and B aligned & x^2=-4 a(-8 a) & x^2=32 a^2 x= ± 4 2 a & A(-4 2 a,-8 a), B(4 2 a,-8 a) & A B^2=(8 2 a)^2=128 a^2=s aligned Length of LR =r=4 a aligned & r s = 4 a 128 a^2 =(1)/(16) & 32 a=16 a=(1)/(2) & 24 a=12 Ans. aligned

JEE Advanced 2024Paper 2

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

JEE Advanced 2024 · Paper 2 · Mathematics · Parabola

A normal with slope 1 6 is drawn from the point (0,-α) to the...

A normal with slope

\frac{1}{\sqrt{6}}

is drawn from the point

(0,-\alpha)

to the parabola

x^2=-4 a y

, where

a>0

. Let

L

be the line passing through

(0,-\alpha)

and parallel to the directrix of the parabola. Suppose that

L

intersects the parabola at two points

A

and

B

. Let

r

denote the length of the latus rectum and

s

denote the square of the length of the line segment

A B

. If

r: s=1: 16

, then the value of 24ais ________. A normal with slope 1 6 is drawn from the point (0,-α) to the parabola x^2=-4 a y , where a>0 . Let L be the line

Correct answer

Step-by-step solution

\frac{d y}{d x}=\left.\frac{x}{-2 a} \Rightarrow \frac{d y}{d x}\right|_N=-t

Slope of normal

=\frac{1}{t}=\frac{1}{\sqrt{6}} \Rightarrow t=\sqrt{6}

Now,

\frac{-\mathrm{at}^2+\alpha}{2 \mathrm{at}}=\frac{1}{\mathrm{t}}\Rightarrow-\mathrm{at}^2+\alpha=2 \mathrm{a} \Rightarrow-6 \mathrm{a}+\alpha=2 \mathrm{a} \Rightarrow \alpha=8 \mathrm{a}

For A and B

\begin{aligned} & x^2=-4 a(-8 a) \\ & \Rightarrow x^2=32 a^2 \Rightarrow x= \pm 4 \sqrt{2} a \\ & \therefore A(-4 \sqrt{2} a,-8 a), B(4 \sqrt{2} a,-8 a) \\ & \therefore A B^2=(8 \sqrt{2} a)^2=128 a^2=s\end{aligned}

\therefore

Length of LR

=\mathrm{r}=4 \mathrm{a}

\begin{aligned} & \Rightarrow \frac{\mathrm{r}}{\mathrm{s}}=\frac{4 \mathrm{a}}{128 \mathrm{a}^2}=\frac{1}{16} \\ & \therefore 32 \mathrm{a}=16 \Rightarrow \mathrm{a}=\frac{1}{2} \\ & \therefore 24 \mathrm{a}=12 \text{ Ans. }\end{aligned}

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