TS EAMCET · Maths · Hyperbola
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 a hyperbola passing through the foci of the given ellipse and \(e_1 e_2=1\), then the equation of such a hyperbola among the following is
- A \(\frac{x^2}{9}-\frac{y^2}{16}=1\)
- B \(\frac{y^2}{9}-\frac{x^2}{16}=1\)
- C \(\frac{x^2}{9}-\frac{y^2}{25}=1\)
- D \(\frac{x^2}{25}-\frac{y^2}{9}=1\)
Answer & Solution
Correct Answer
(B) \(\frac{y^2}{9}-\frac{x^2}{16}=1\)
Step-by-step Solution
Detailed explanation
Equation of ellipse \(\frac{x^2}{16}+\frac{y^2}{25}=1\) Foci \(\quad=(0, \pm 3)\) \(\Rightarrow \quad e_2=\sqrt{1-\frac{16}{25}}=\frac{3}{5}\) given \(e_2 e_2=1\) (where \(e_2\) is eccentricity of hyperbola) Let equation of hyperbola \(-\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) its…
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