JEE Main · 2004 · 24 Apr 2004 · Offline · Parabola

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a \neq = 0

and the line

2 b x + 3 cy + 4 d = 0

passes through the points of intersection of the parabolas

y^{2} = 4 a x

and

x^{2} = 4 a y

, then

(A) $d^{2} + (2 b + 3 c)^{2} = 0$ ✓ Correct
(B) $d^{2} + (3 b + 2 c)^{2} = 0$
(C) $d^{2} + (2 b - 3 c)^{2} = 0$
(D) $d^{2} + (3 b - 2 c)^{2} = 0$

Answer: (A)

Solution

Find where

y^{2} = 4 a x

and

x^{2} = 4 a y

meet: substituting

x = \frac{y ^{2}}{4 a}

into

x^{2} = 4 a y

gives

\frac{y ^{4}}{16 a ^{2}} = 4 a y

, so

y^{4} = 64 a^{3} y

, i.e.

y = 0

y = 4 a

. The intersection points are

(0, 0)

and

(4 a, 4 a)

. For

2 b x + 3 cy + 4 d = 0

to pass through

(0, 0)

we need

d = 0

, and through

(4 a, 4 a)

we need

8 ab + 12 a c = 0

, i.e.

2 b + 3 c = 0

. Both conditions together say

d^{2} + (2 b + 3 c)^{2} = 0

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option

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