• # question_answer If $a\ne 0\And$ the line $2bx+3cy+4d=0$ passes through the point of intersection of parabola ${{y}^{2}}=4ax$ and ${{x}^{2}}=4ay,$then A) ${{d}^{2}}+{{(3b-2c)}^{2}}=0$  B) ${{d}^{2}}+{{(3b+2c)}^{2}}=0$ C) ${{d}^{2}}+{{(2b-3c)}^{2}}=0$  D) ${{d}^{2}}+{{(2b+3c)}^{2}}=0$

Two given parabola intersect at (0,0) and (4a, 4a). So, the common chord is y = x. On comparing y = x with given line, we get $\frac{2b}{1}=\frac{3c}{-1}=\frac{4d}{0}$$\Rightarrow$d = 0 and $2b=3c=0$