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 $15$

Given two conics:

$C_1:x^2+y^2=2$

$C_2:y^2=8x$

Lets assume common tangents touches circle at $P(x_1,y_1)$ and $R(x_2,y_2)$

Then,

$C_1(x_1,y_1):x_1^2+y_1^2-2=0$

$C_2(x_2,y_2):y_2^2-8x_2=0$

The tangent space to the conics is

$m_1=\frac{d{y}_1}{d{x}_1}=-\frac{x_1}{y_1}$ … (i)

$m_2=\frac{d{y}_2}{d{x}_2}=\frac{4}{y_2}$ …(ii)

The tangency condition reads:

$\{\begin{array}{c}{y}_2-y_1=m_1(x_2-x_1)\dots \left(iii\right)\\ y_1-y_2=m_2(x_1-x_2)\dots \left(iv\right)\end{array}$

On solving these four eqs, we get

$x_1=-1,x_2=2,y_1=\pm 1,y_2=\pm 4$

Therefore,

$P\equiv (-1,1)$

$Q\equiv (-1,-1)$

$R\equiv (2,4)$

$S\equiv (2,-4)$

Now the area of the quadrilateral $PQRS$:

$\frac{1}{2}\times (PQ+RS)\times \left(\text{Distance between PQ and RS}\right)$

$=\frac{1}{2}\times (2+8)\times (1+2)$

$=\frac{1}{2}\times 10\times 3=15$ sq units.