Question

Consider the parabola $y^2=8x$. Let ${\Delta }_1$ be the area of the triangle formed by the end points of its latus rectum and the point $P(\frac{1}{2},2)$ on the parabola, and ${\Delta }_2$ be the area of the triangle formed by drawing tangents at $P$ and at the end points of the latus rectum. Then $\frac{{\Delta }_1}{{\Delta }_2}$ is

Answer 1

Clearly, $ar\left(\Delta ABC\right)=2\left(ar\Delta PQR\right)$
${\Delta }_1=2{\Delta }_2$
$\Rightarrow \frac{{\Delta }_1}{{\Delta }_2}=2$