Question

Consider the circle $x^2+y^2=9$ and the parabola $y^2=8x$. They intersect at $P$ and $Q$ in the first and the fourth quadrants, respectively. Tangents to the circle at $P$ and $Q$ intersect the $x$-axis at $R$ and tangents to the parabola at $P$ and $Q$ intersect the x-axis at $S$. The ratio of the areas of the triangles $PQS$ and $PQR$ is

• A. $1:\sqrt{2}$
• B. $1:2$
• C. $1:4$
• D. $1:8$

Answer 1

The correct option is C $1:4$

$S_1:x^2+y^2=9$

$S_2:y^2=8x$

$\therefore x^2+8x-9=0$

$\therefore (x+9)(x-1)=0$

$\therefore x=1$ ....[$x=-9$ is not a solution because for the parabola $x>0$]

$y^2=8x=8$

$⟹y=\pm 2\sqrt{2}$

$P\equiv (1,2\sqrt{2})$

$Q\equiv (1,-2\sqrt{2})$

Tangent to $S_1=0$ at $(1,2\sqrt{2})$ is $T_1=0$

$x+2\sqrt{2}y=9$

Tangent to $S_1=0$ at $(1,-2\sqrt{2})$ is $T_1=0$

$x-2\sqrt{2}y=9$

They intersects the x-axis at $(9,0)$

$⟹R\equiv (9,0)$

Tangent to $S_2=0$ at $(1,2\sqrt{2})$ is $T_2=0$

$2\sqrt{2}y=4(x+1)$

Tangent to $S_2=0$ at $(1,-2\sqrt{2})$ is $T_2=0$

$-2\sqrt{2}y=4(x+1)$

They intersects the x-axis at $(-1,0)$

$⟹S\equiv (-1,0)$

Let the line $PQ$ intersect x-axis at $M$

Co-ordinates of $M=(1,0)$

$\frac{A\left(\Delta PQS\right)}{A\left(\Delta PQR\right)}=\frac{\frac{1}{2}\times PQ\times SM}{\frac{1}{2}\times PQ\times RM}$

$=\frac{SM}{RM}$

$=\frac{1-(-1)}{9-1}=\frac{1}{4}$

The answer is option (C).