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TS 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

  1. A \(\frac{x^2}{144}-\frac{y^2}{9}=1\)
  2. B \(\frac{x^2}{169}-\frac{y^2}{25}=1\)
  3. C \(\frac{x^2}{144}-\frac{y^2}{25}=1\)
  4. D \(\frac{x^2}{25}-\frac{y^2}{9}=1\)
Verified Solution

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,…