Question:
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 $e_1e_2=1$ then equation of the hyperbola is
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 $e_1e_2=1$ then equation of the hyperbola is
Updated On: Jun 14, 2022
- $\frac {x^2}{9}-\frac {y^2}{16}=1$
- $\frac {x^2}{16}-\frac {y^2}{9}=-1$
- $\frac {x^2}{9}-\frac {y^2}{25}=1$
- None of these
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The Correct Option is B
Solution and Explanation
The eccentricity of $\frac {x^2}{16}+\frac {y^2}{25}=1$ is
$e_1=(\sqrt{1-\frac{16}{25}})=\frac{3}{5}$
$\therefore e_2=\frac{5}{3}$ $[\because e_1e_2=1]$
$\Rightarrow$ Foci of ellipse $(0,\pm 3)$
$\Rightarrow$ Equation of hyperbola is $\frac{x^2}{16}-\frac{y^2}{9}=-1.$
$e_1=(\sqrt{1-\frac{16}{25}})=\frac{3}{5}$
$\therefore e_2=\frac{5}{3}$ $[\because e_1e_2=1]$
$\Rightarrow$ Foci of ellipse $(0,\pm 3)$
$\Rightarrow$ Equation of hyperbola is $\frac{x^2}{16}-\frac{y^2}{9}=-1.$
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