Question

If $a\ne 0$ and the line $2bx+3cy+4d=0$ passe 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

The correct option is D $d^2+(2b+3c{)}^2=0$

$y^2=4ax$ and $x^2=4ay$ intersect each other at $(0,0)$ and $(4a,4a)$

putting $(0,0)$ in the equation of line we get $d=0$ . .....(1)

again putting $(4a,4a)$ in the equation we get

$8ba+12ca=0$

$\Rightarrow$$2b+3c=0$ ..........(2)

from (1) and (2) $d^2+{(2b+3c)}^2=0$

so option D is correct