Question

The common tangents to the circle $x^2+y^2=2$ and the parabola $y^2=8x$ touch the circle at the points $P,Q$ and the parabola at the points $R,S$. Then the area of the quadrilateral $PQRS$ is

• A. $3$
• B. $6$
• C. $9$
• D. $15$

Answer 1

The correct option is D $15$


Let the tangent to the parabola be,
$y=mx+\frac{2}{m}$
This is tangent to the circle, so distance from origin is equal to radius of the circle,
$\frac{|2/m|}{\sqrt{1+m^2}}=\sqrt{2}\Rightarrow \frac{4}{m^2}=2(1+m^2)\Rightarrow m^4+m^2-2=0\Rightarrow (m^2+2)(m^2-1)=0\Rightarrow m=\pm 1$

Therefore, the tangents are,
$y=x+2y=-x-2\Rightarrow x+2=-x-2\Rightarrow x=-2,y=0$

$\therefore$ Point $T=(-2,0)$

Equation of chord of contact $PQ$ is
$T=0\Rightarrow x(-2)+y\left(0\right)-2=0\Rightarrow x=-1$
Point $P=(-1,1);Q=(-1,-1)$

Equation of chord of contact $RS$ is
$T=0\Rightarrow y\left(0\right)-4(x-2)=0\Rightarrow x=2$
Point $R=(2,4);S=(2,-4)$

Area of the quadrilateral is,
$=2[\text{Area of}△TVR-\text{Area of}△TUP]=2[\frac{1}{2}\times TV\times RV-\frac{1}{2}\times TU\times PU]=2[\frac{1}{2}\times 4\times 4-\frac{1}{2}\times 1\times 1]=15\text{sq. units}$