Question

From a point $A$ common tangents are drawn to the circle $x^2+y^2=a^2/2$ and parabola $y^2=4ax$. The area of the quadrilateral found of the common tangents to the chord of contact of the parabola is

• A. $\frac{15a^3}{4}$ sq. units
• B. $\frac{15a^2}{4}$ sq. units
• C. $\frac{7a^2}{4}$ sq. units
• D. $\frac{a^2}{4}$ sq. units

Answer 1

The correct option is B $\frac{15a^2}{4}$ sq. units

Equation of any tangent of the parabola, $y^2=ax$ is $y=mx+\frac{a}{m}$.

This line touch the circle $x^2+y^2=\frac{a^2}{2}$.
Therefore, $BC=2\sqrt{O{B}^2-O{K}^2}$
$=2\sqrt{\frac{a^2}{2}-\frac{a^2}{4}}$
$=2\sqrt{\frac{a^2}{4}}=a$
and we know that $DE$ is the latusrectum of the parabola, so its length is $4a$.
Thus, area of trapezium $BCDE$
$=\frac{1}{2}(BC+DE)\left(KL\right)$
$=\frac{1}{2}(a+4a)\left[\frac{3a}{2}\right]$
$=\frac{15a^2}{4}$ sq units