Hyperbola Ques 30
If a hyperbola passes through the point $P(\sqrt{2}, \sqrt{3})$ and has foci at $( \pm 2,0)$, then the tangent to this hyperbola at $P$ also passes through the point
(2017 Main)
(a) $(3 \sqrt{2}, 2 \sqrt{3})$
(b) $(2 \sqrt{2}, 3 \sqrt{3})$
(c) $(\sqrt{3}, \sqrt{2})$
(d) $(-\sqrt{2},-\sqrt{3})$
Show Answer
Answer:
Correct Answer: 30.(b)
Solution:
Formula:
- Let the equation of hyperbola be $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$.
$ \begin{array}{lc} \therefore & a e=2 \Rightarrow a^{2} e^{2}=4 \\ \Rightarrow & a^{2}+b^{2}=4 \Rightarrow b^{2}=4-a^{2} \\ \therefore & \frac{x^{2}}{a^{2}}-\frac{y^{2}}{4-a^{2}}=1 \end{array} $
Since, $(\sqrt{2}, \sqrt{3})$ lie on hyperbola.
$ \begin{aligned} & \therefore \quad \frac{2}{a^{2}}-\frac{3}{4-a^{2}}=1 \\ & \Rightarrow \quad 8-2 a^{2}-3 a^{2}=a^{2}\left(4-a^{2}\right) \\ & \Rightarrow \quad 8-5 a^{2}=4 a^{2}-a^{4} \\ & \Rightarrow \quad a^{4}-9 a^{2}+8=0 \\ & \Rightarrow \quad\left(a^{4}-8\right)\left(a^{4}-1\right)=0 \Rightarrow a^{4}=8, a^{4}=1 \\ & \therefore \quad a=1 \end{aligned} $
Now, equation of hyperbola is $\frac{x^{2}}{1}-\frac{y^{2}}{3}=1$.
$\therefore$ Equation of tangent at $(\sqrt{2}, \sqrt{3})$ is given by
$ \sqrt{2} x-\frac{\sqrt{3} y}{3}=1 \Rightarrow \sqrt{2} x-\frac{y}{\sqrt{3}}=1 $
which passes through the point $(2 \sqrt{2}, 3 \sqrt{3})$.
BETA
