JEE MAIN - Mathematics (2004 - No. 46)
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 :
$${d^2} + {\left( {3b - 2c} \right)^2} = 0$$
$${d^2} + {\left( {3b + 2c} \right)^2} = 0$$
$${d^2} + {\left( {2b - 3c} \right)^2} = 0$$
$${d^2} + {\left( {2b + 3c} \right)^2} = 0$$
Explanation
Solving equations of parabolas
$${y^2} = 4ax$$ and $${x^2} = 4ay$$
we get $$(0,0)$$ and $$(4a, 4a)$$
Substituting in the given equation of line
$$2bx+3cy+4d=0,$$
we get $$d=0$$
and $$2b+3c=0$$ $$ \Rightarrow {d^2} + {\left( {2b + 3c} \right)^2} = 0$$
$${y^2} = 4ax$$ and $${x^2} = 4ay$$
we get $$(0,0)$$ and $$(4a, 4a)$$
Substituting in the given equation of line
$$2bx+3cy+4d=0,$$
we get $$d=0$$
and $$2b+3c=0$$ $$ \Rightarrow {d^2} + {\left( {2b + 3c} \right)^2} = 0$$
Comments (0)
