Question

Let $P$ be the point of intersection of the common tangents to the parabola $y^2=12x$ and the hyperbola $8x^2-y^2=8.$ If $S$ and $S^{\prime }$ denote the foci of the hyperbola where $S$ lies on the positive $x-$axis then $P$ divides $S{S}^{\prime }$ in a ratio

• A. $14:13$
• B. $13:11$
• C. $5:4$
• D. $2:1$

Answer 1

The correct option is C $5:4$

Equation of parabola $y^2=12x$
So equation of its tangent : $y=3x+\frac{3}{m}$
Equation of hyperbola $\frac{x^2}{1}-\frac{y^2}{8}=1$
Eccentricity of hyperbola $e=\sqrt{1+8}=3$
$S(ae,0)=S(3,0)\&S^{\prime }(-ae,0)=S^{\prime }(-3,0)$

And equation of its tangent : $y=mx\pm \sqrt{m^2-8}$
Both tangent are coomon tangents
Therefore, $\frac{9}{m^2}=m^2-8$
Let $m^2=t$
$t^2-8t-9=0$
$\Rightarrow t=m^2=9,-1\left(\text{not possible}\right)$
$\Rightarrow m=\pm 3$

$\therefore \begin{array}{c}y=3x+1\\ y=-3x-1\end{array}$
Therefore point of intersection of common tangents $P(-\frac{1}{3},0)$
Let $P$ divides $S{S}^{\prime }$ in a ratio of $m:n$
$\therefore P(-\frac{1}{3},0)=P(\frac{-3m+3n}{m+n},0)$
$\Rightarrow -m-n=-9m+9n\Rightarrow \frac{m}{n}=\frac{5}{4}$