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 equation of parabolas are $y^2=4ax$ and $x^2=4ay$

On solving these we get $x=0$ and $x=4a$

Also $y=0$ and $y=4a$

Therefore point of intersection of parabolas are $A(0,0)$ and $B(4a,4a)$

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

$\therefore d=0$ ...(1)

$2b.4a+3c.4a+4d=0\Rightarrow 2ab+3ac+d=0$

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

On squaring (1) and (2) and then adding, we get

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