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

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

The points of intersection are (0, 0), (4a, 4a)
(0, 0) lies on $2bx+3cy+4d=0\Rightarrow d=0$
(4a, 4a) lies on $2bx+3cy+4d=0\Rightarrow 8ab+12ac=0\Rightarrow 2b+3c=0[\because a\ne 0]$
$\therefore d^2+(2b+3c{)}^2=0$