JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Parabola

A normal with slope

\frac{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 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

24 a

is ____.

Answer: 12

Solution

For the parabola

x^{2} = - 4 a y

, a point on it is

(2 a t, - a t^{2})

and the slope of the normal is

t

.

Given the normal has slope

\frac{1}{6}

, so

t = 6

.

The normal passes through

(0, - α)

\frac{- a t ^{2} + α}{2 a t} = \frac{1}{t}

- a t^{2} + α = 2 a

α = 2 a + a t^{2} = 2 a + 6 a = 8 a

The line

L

y = - 8 a

. Substituting into the parabola:

x^{2} = - 4 a (- 8 a) = 32 a^{2} \Rightarrow x = \pm 42 a

A = (- 42 a, - 8 a)

and

B = (42 a, - 8 a)

.

Length of

A B

∣ A B ∣ = 82 a

, so

s = (82 a)^{2} = 128 a^{2}

.

Length of latus rectum:

r = 4 a

.

Given

\frac{r}{s} = \frac{1}{16}

\frac{4 a}{128 a ^{2}} = \frac{1}{16} \Rightarrow \frac{1}{32 a} = \frac{1}{16} \Rightarrow a = \frac{1}{2}

Therefore:

24 a = 12

JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Integer answer