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 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 $y=4a.$ $\therefore$ The point of intersection of parabolas are A(0,0) and $B(4a,4a)$ . Also line $2bx+3cy+4d=0$ passes through A and B. $\therefore$ $d=0$ ...(i) and $2b.4a+3c-4a+4d=0$ $\Rightarrow$ $2ab+3ac+d=0$ $\Rightarrow$ $a(2b+3c)=0$ $(\because d=0)$ $\Rightarrow$ $2b+3c=0$ ...(ii) On squaring Eqs. (i) and (ii) and then adding, we get $d^2+{(2b+3c)}^2=0$