Question

If $a\ne 0$ and the line $2bx+3cy+4d=0$ passes through the point of intersection of the parabola's $y^2=4ax$ and $x^2=4ay$, then

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

Answer 1

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

The equation of the parabolas are $y^2=4ax,x^2=4ay$

on solving these we get $x=0,x=4a$ also $y=0,y=4a$

therefore the point of intersection of parabolas are $A(0,0),B(4a,4a)$

also line $2bx+3cy+4d=0$ passes through A and B

$d=0........\left(1\right)$ or $2ab+3ac+d=0$

$a(2b+3c)=0.......(d=0)$

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

On squaring equation 1 and 2 and then adding we get $d^2+(2b+3c{)}^2=0$