JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Parabola

For the parabola $y^2=8x$, we have $4a=8$, so $a=2$. The focus is at $(2,0)$ and the latus rectum endpoints are $L(2,4)$ and $M(2,-4)$.

The point $P(\frac{1}{2},2)$ lies on the parabola (verify: $2^2=4=8\cdot \frac{1}{2}$).

${\Delta }_1$ is the area of triangle $LPM$ with vertices $(2,4),(\frac{1}{2},2),(2,-4)$.

Using the formula: ${\Delta }_1=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$
$=\frac{1}{2}|2(2-(-4))+\frac{1}{2}((-4)-4)+2(4-2)|=\frac{1}{2}|12-4+4|=6$.

${\Delta }_2$ is the area of the triangle formed by the three tangents. By a standard property of parabolas, when tangents are drawn at the endpoints of the latus rectum and at a point on the parabola, the ratio $\frac{{\Delta }_1}{{\Delta }_2}=2$.