AP EAMCET · Maths · Hyperbola
A hyperbola passing through a focus of the ellipse \(\frac{x^2}{169}+\frac{y^2}{25}=1\). Its transverse and conjugate axes coincide respectively with the major and minor axes of the ellipse. The product of eccentricities is 1 . Then, the equation of the hyperbola is
- A \(\frac{x^2}{144}-\frac{y^2}{9}=1\)
- B \(\frac{x^2}{169}-\frac{y^2}{25}=1\)
- C \(\frac{x^2}{144}-\frac{y^2}{25}=1\)
- D \(\frac{x^2}{25}-\frac{y^2}{9}=1\)
Answer & Solution
Correct Answer
(C) \(\frac{x^2}{144}-\frac{y^2}{25}=1\)
Step-by-step Solution
Detailed explanation
Let the equation of hyperbola be \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \] Given equation of ellipse is \[ \frac{x^2}{(13)^2}+\frac{y^2}{(5)^2}=1 \] Here,…
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