Question

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

• A. $\frac{15a^2}{4}$
• B. $\frac{17a^2}{4}$
• C. $\frac{19a^2}{4}$
• D. $\frac{21a^2}{4}$

Answer 1

The correct option is A $\frac{15a^2}{4}$

Equation of any tangent to the parabola $y^2=4ax$ is $y=mx+\frac{a}{m}$

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

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

$\Rightarrow m^4+m^2-2\Rightarrow (m^2+2)(m^2-1)=0$

As $m^2+2>0,m^2-1=0\Rightarrow m=\pm 1$

Thus, the two common tangents are $y=x+a$ and $y=-x-a$

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

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

$(-a)x+0\left(y\right)=\frac{a^2}{2}\Rightarrow x=-\frac{a}{2}$

and chord of contact of $A(-a,0)$ for the parabola $y^2=4ax$ is

$\left(0\right)y=2a(x-a)\Rightarrow x=a$

Length of $BC=2\sqrt{{OB}^2-{OK}^2}=2\sqrt{\frac{a^2}{2}-\frac{a^2}{4}}=2\sqrt{\frac{a^2}{4}}=a$

Note that $DE$ is the latus rectum of the parabola, so its length is $4a$.

Thus area of the trapezium $BCDE=\frac{1}{2}(BC+DE)\left(KL\right)$

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