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

• A

  $\frac{15 a^3}{4}$ sq units
• Correct answer
  B

  $\frac{15 a^2}{4}$ sq units
• C

  $\frac{7 a^2}{4}$ sq units
• D

  $\frac{a^2}{4}$ sq unit

Solution

Equation of any tangent of the parabola, $y^2=4 a x$ is $y=m x+\frac{a}{m}$. This line will touch the circle $x^2+y^2=\frac{a^2}{2}$.
$\begin{aligned} \therefore \quad B C & =2 \sqrt{O B^2-O K^2} \\ & =2 \sqrt{\frac{a^2}{2}-\frac{a^2}{4}} \\ & =2 \sqrt{\frac{a^2}{4}}=a \end{aligned}$ and we know that $D E$ is the latusrectum of the parabola, so its length is $4 a$. Thus, area of trapezium $B C D E$ $\begin{aligned} & =\frac{1}{2}(B C+D E)(K L) \\ & =\frac{1}{2}(a+4 a)\left[\frac{3 a}{2}\right] \\ & =\frac{15 a^2}{4} \text { squnits } \end{aligned}$

Hence, the correct answer is option: 'B'.