Hyperbola 1 Question 14
15.
Let the eccentricity of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ be reciprocal to that of the ellipse $x^{2}+4 y^{2}=4$. If the hyperbola passes through a focus of the ellipse, then
(a) the equation of the hyperbola is $\frac{x^{2}}{3}-\frac{y^{2}}{2}=1$
(b) a focus of the hyperbola is $(2,0)$
(c) the eccentricity of the hyperbola is $\sqrt{\frac{5}{3}}$
(d) the equation of the hyperbola is $x^{2}-3 y^{2}=3$
(2011)
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Answer:
Correct Answer: 15. (b, d)
Solution:
- Here, equation of ellipse is $\frac{x^{2}}{4}+\frac{y^{2}}{1}=1$
$\Rightarrow \quad e^{2}=1-\frac{b^{2}}{a^{2}}=1-\frac{1}{4}=\frac{3}{4}$
$\therefore \quad e=\frac{\sqrt{3}}{2}$ and focus $( \pm a e, 0) \Rightarrow( \pm \sqrt{3}, 0)$
For hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1, e _1^{2}=1+\frac{b^{2}}{a^{2}}$
where,
$ e _1^{2}=\frac{1}{e^{2}}=\frac{4}{3} \Rightarrow 1+\frac{b^{2}}{a^{2}}=\frac{4}{3} $
$ \therefore \quad \frac{b^{2}}{a^{2}}=\frac{1}{3} $
and hyperbola passes through $( \pm \sqrt{3}, 0)$
$\Rightarrow \quad \frac{3}{a^{2}}=1 \quad \Rightarrow \quad a^{2}=3$
From Eqs. (i) and (ii), $b^{2}=1$
$\therefore$ Equation of hyperbola is $\frac{x^{2}}{3}-\frac{y^{2}}{1}=1$
Focus is $\left( \pm a e _1, 0\right) \Rightarrow \pm \sqrt{3} \cdot \frac{2}{\sqrt{3}}, 0 \Rightarrow( \pm 2,0)$
Hence, (b) and (d) are correct answers.
