Question

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

Answer 1

Answer: (A)

The given parabolas are $y^2=4ax$ and $x^2=4ay$.
Solving these, we get $A(0,0),B(4a,4a)$ as their points of intersection
Also the line $2bx+3cy+4d=0$ passes through $A$ and $B$.
$\therefore d=0$ and $2b\cdot 4a+3c\cdot 4a=0$
$\Rightarrow a\left(2b+3c\right)=0$
$\Rightarrow 2b+3c=0\left[\because a\ne 0\right]$
$\therefore d^2+{\left(2b+3c\right)}^2=0$
$=0+0=0$