Hyperbola Ques 4
- If $e _1$ is the eccentricity of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{25}=1$ and $e _2$ is the eccentricity of the hyperbola passing through the foci of the ellipse and $e _1 e _2=1$, then equation of the hyperbola is
(a) $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$
(b) $\frac{x^{2}}{16}-\frac{y^{2}}{9}=-1$
(c) $\frac{x^{2}}{9}-\frac{y^{2}}{25}=1$
(d) None of these
Show Answer
Answer:
Correct Answer: 4.(b)
Solution:
- The eccentricity of $\frac{x^{2}}{16}+\frac{y^{2}}{25}=1$ is
$$ \begin{array}{rlrl} e _1 & =\sqrt{1-\frac{16}{25}}=\frac{3}{5} \\ \therefore \quad e _2 & =\frac{5}{3} & {\left[\because e _1 e _2=1\right]} \end{array} $$
$\Rightarrow$ Foci of ellipse $(0, \pm 3)$
$\Rightarrow$ Equation of hyperbola is $\frac{x^{2}}{16}-\frac{y^{2}}{9}=-1$.
BETA
