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.
$${d^2} + {\left( {3b - 2c} \right)^2} = 0$$
B.
$${d^2} + {\left( {3b + 2c} \right)^2} = 0$$
C.
$${d^2} + {\left( {2b - 3c} \right)^2} = 0$$
D.
$${d^2} + {\left( {2b + 3c} \right)^2} = 0$$
Answer :
$${d^2} + {\left( {2b + 3c} \right)^2} = 0$$
Solution :
Solving equations of parabolas
$${y^2} = 4ax$$ and $${x^2} = 4ay$$
we get $$\left( {0,\,0} \right)$$ and $$\left( {4a,\,4a} \right)$$
Substituting in the given equation of line
$$2bx + 3cy + 4d = 0,$$
we get $$d=0$$ and $$2b + 3c = 0$$
$$ \Rightarrow {d^2} + {\left( {2b + 3c} \right)^2} = 0$$