Hyperbola Ques 1
- Let $P$ be the point of intersection of the common tangents to the parabola $y^2=12 x$ and the hyperbola $8 x^2-y^2=8$. If $S$ and $S^{\prime}$ denotes the foci of the hyperbola where $S$ lies on the positive $X$-axis then $P$ divides $S S^{\prime}$ in a ratio
(2019 Main, 12 April I)
(a) $13: 11$
(b) $14: 13$
(c) $5: 4$
(d) $2: 1$
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Answer:
Correct Answer: 1.(c)
Solution: (c) Equation of given parabola $y^2=12 x$ $\quad$ ……..(i)
and hyperbola $8 x^2-y^2=8$ $\quad$ ……..(ii)
Now, equation of tangent to parabola $y^2=12 x$ having slope ’ $m$ ’ is $y=m x+\frac{3}{m}$ $\quad$ ……..(iii)
and equation of tangent to hyperbola
$\frac{x^2}{1}-\frac{y^2}{8}=1$ having slope ’ $m$ ’ is
$ y=m x \pm \sqrt{1^2 m^2-8} $ $\quad$ ……..(iv)
Since, tangents (iii) and (iv) represent the same line
$\therefore m^2-8=\left(\frac{3}{m}\right)^2 $
$\Rightarrow m^4-8 m^2-9=0 $
$\Rightarrow \left(m^2-9\right)\left(m^2+1\right)=0 $
$\Rightarrow m= \pm 3 .$
Now, equation of common tangents to the parabola (i) and hyperbola (ii) are $y=3 x+1$ and $y=-3 x-1$
$\because \quad $ Point ’ $P$ is point of intersection of above common tangents,
$ \therefore \quad P(-1 / 3,0) $
and focus of hyperbola $S(3,0)$ and $S^{\prime}(-3,0)$.
Thus, the required ratio $=\frac{P S}{P S^{\prime}}=\frac{3+1 / 3}{3-1 / 3}=\frac{10}{8}=\frac{5}{4}$
BETA
