Question

Find the point on the parabola $y^2=4ax\left(a>0\right)$ which forms a triangle of area $3a^2$ with the vertex and focus of the parabola.

Answer 1

Vertex $(0,0)$ Focus $(a,0)$ Point $P(x,y)$

Area of triangle $=\frac{1}{2}\times$ Base $\times$ Height

$3a^2=\frac{1}{2}\times a\times y$

$y=6a$

$(6a{)}^2=4ax$

$x=9a$

$P(x,y)=P(9a,6a)$