JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Parabola

Let P be a point on the parabola

y^{2} = 4 ax

, where

a > 0

. The normal to the parabola at P meets the x-axis at a point Q. The area of the triangle PFQ, where F is the focus of the parabola, is 120. If the slope m of the normal and a are both positive integers, then the pair (a,m) is

(A) $(2, 3)$ ✓ Correct
(B) $(1, 3)$
(C) $(2, 4)$
(D) $(3, 4)$

Answer: (A)

Solution

Let

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

be a point on the parabola

y^{2} = 4 a x

.

The normal at P has equation:

y = - t x + 2 a t + a t^{3}

When

y = 0

x = 2 a + a t^{2}

, so

Q (2 a + a t^{2}, 0)

The focus is

F (a, 0)

.

The area of triangle PFQ is:

Area = \frac{1}{2} (a + a t^{2}) (2 a t) = \frac{1}{2} \cdot a (1 + t^{2}) \cdot 2 a t = a^{2} t (1 + t^{2})

Given:

a^{2} t (1 + t^{2}) = 120

Since the slope of the normal is

m = - t

, we have

t = - m

(but we take

t

positive for the calculation).

So:

a^{2} (1 + m^{2}) m = 120

Trying

(a, m) = (2, 3)

4 (1 + 9) \cdot 3 = 4 \cdot 10 \cdot 3 = 120

✓

Answer: (A)

JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Single correct option

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JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Parabola

Solution

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