Question

If $a \neq 0$and the line $2bx + 3cy + 4d = 0$passes through the points of intersection of the parabolas $y^{2} = 4ax$and $x^{2} = 4ay$then

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

Answer 1

Answer: (D)

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

Answer 2

Given parabolas are $y^{2} = 4ax$ .....(i) and $x^{2} = 4ay$

.....(ii)

from (i) and (ii) $\left( \frac{x^{2}}{4a} \right)^{2} = 4ax$⇒ $x^{4} - 64a^{3}x = 0$ ⇒ $x = 0$, $4a$∴$y = 0,4a$

So points of intersection are $(0,0)$ and $(4a,4a)$

Given, the line $2bx + 3cy + 4d = 0$passes through (0,0) and $(4a,4a)$

∴$d = 0$⇒ $d^{2} = 0$ and $(2b + 3c)^{2} = 0$ $(\because a \neq 0)$

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