JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Parabola

Let P be a point on the parabola,

y^{2} = 12 x

and N be the foot of the perpendicular drawn from

P

on the axis of the parabola. A line is now drawn through the mid-point

M

P N

, parallel to its axis which meets the parabola at

Q

. If the

y

-intercept of the line

N Q

\frac{4}{3}

, then:

(A) $MQ = \frac{1}{3}$
(B) $MQ = \frac{1}{4}$ ✓ Correct
(C) $PN = 4$
(D) $PN = 3$

Answer: (B)

Solution

Take

P = (3 t^{2}, 6 t)

y^{2} = 12 x

; then

N = (3 t^{2}, 0)

is its foot on the axis and the midpoint of

P N

M = (3 t^{2}, 3 t)

. The horizontal line

y = 3 t

meets the parabola again at

Q = (\frac{3 t ^{2}}{4}, 3 t)

. The line

N Q

has

y

-intercept

4 t

, and setting this equal to

\frac{4}{3}

gives

t = \frac{1}{3}

. Then

M Q = 3 t^{2} - \frac{3 t ^{2}}{4} = \frac{9 t ^{2}}{4} = \frac{1}{4}

JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Single correct option

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