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From a point A common tangents are drawn to the circle $x^2+y^2=\frac{a^2}{2}$ and parabola $y^2=4 a x$ . Find the area of the quadrilateral formed by the common tangents, the chord of contact of the circle and the chord of contact of the parabola.

Option: 1
$\begin{aligned} & \frac{25 a^3}{4} \\ \end{aligned}$

Option: 2
$\frac{5 a}{7} \\$

Option: 3
$\frac{15 a^2}{4} \\$

Option: 4
$-\frac{13 a^2}{9}$

Answers (1)

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Equation of any tangent to the parabola, $y^2=4 a x \text { is } y=m x+\frac{a}{m} \text {. }$ .

This line will touch the circle $x^2+y^2=\frac{a^2}{2} .$

$\begin{aligned} & \text { If, }\left(\frac{a}{m}\right)^2=\frac{a^2}{2}\left(m^2+1\right) \\ & \Rightarrow \frac{1}{m^2}=\frac{1}{2}\left(m^2+1\right) \\ & \Rightarrow 2=m^4+m^2 \\ & \Rightarrow m^4+m^2-2=0 \\ & \Rightarrow\left(m^2-1\right)\left(m^2+2\right)=0 \\ & \Rightarrow m^2-1=0, m^2=-2 \\ & \Rightarrow m= \pm 1 \end{aligned}$

Therefore, two common tangents are $y=x+a \text { and } y=-x-a \text {. }$

These two intersect at A(-a ,0 ).

The chord of contact of A (a ,-0 ) for the circle $x^2+y^2=\frac{a^2}{2}$

$\Rightarrow x=-\frac{a}{2}$

and chord of contact of A(a , -0) for the parabola $y^2=4 a x \text { is } 0$

$\begin{aligned} y & =2 a(x-a) \\ \Rightarrow x & =a \end{aligned}$

Again, length of BC =

$\begin{aligned} & \quad=2 B K \\ & =2 \sqrt{O B^2-O K^2} \\ & =2 \sqrt{\frac{a^2}{2}-\frac{a^2}{4}}=2 \sqrt{\frac{a^2}{4}}=a \end{aligned}$

and we know that, DE is the latus rectum of the parabola, so its length is 4a.

Thus, area of the quadrilateral BCDE =

$\begin{array}{r} =\frac{1}{2}(B C+D E)(K L) \\ \end{array}$

$=\frac{1}{2}(a+4 a)\left(\frac{3 a}{2}\right)=\frac{15 a^2}{4}$

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