Consider the parabola $y^2=8 x$. Let $\Delta_1$ be the area of the triangle formed by the end points of its…
Consider the parabola $y^2=8 x$. Let $\Delta_1$ be the area of the triangle formed by the end points of its latus rectum and the point $P\left(\frac{1}{2}, 2\right)$ 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
Solution
As, we know area of triangle formed by three points on parabola is twice the area of triangle formed by corresponding tangents, i.e. area of $\triangle P Q R=2$ area of $\Delta T_1 T_2 T_3$.
$\therefore \quad \Delta_1=2 \Delta_2$ or $\frac{\Delta_1}{\Delta_2}=2$
