JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Parabola

Let

P

be a variable point on the parabola

y = 4 x^{2} + 1

. Then, the locus of the mid-point of the point

P

and the foot of the perpendicular drawn from the point

P

to the line

y = x

is :

Answer: (C)

Solution

Take

P = (p, 4 p^{2} + 1)

. The foot of the perpendicular from

P

y = x

(s, s)

with

s = \frac{p + 4 p ^{2} + 1}{2}

, and the midpoint

(h, k)

P

and

(s, s)

h = \frac{p + s}{2} = \frac{4 p ^{2} + 3 p + 1}{4}, k = \frac{4 p ^{2} + 1 + s}{2} = \frac{12 p ^{2} + p + 3}{4}

Taking two combinations that separate

p

from

p^{2}

3 h - k = 2 p, h - 3 k = - 8 p^{2} - 2

Eliminating

p

gives

h - 3 k + 2 = - 8 p^{2} = - 2 (3 h - k)^{2}

, so the locus is

2 (3 x - y)^{2} + (x - 3 y) + 2 = 0

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Single correct option