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AP EAMCET · Maths · Hyperbola

The foci of the ellipse \(\frac{x^2}{16}+\frac{y^2}{b^2}=1\) and the hyperbola \(\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}\) coincide. Then, the value of \(b^2\) is

  1. A \(5\)
  2. B \(7\)
  3. C \(9\)
  4. D \(1\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(7\)

Step-by-step Solution

Detailed explanation

Equation of ellipse, \(\frac{x^2}{16}+\frac{y^2}{b^2}=1\) Eccentricity, \(\therefore \quad e=\sqrt{1-\frac{b^2}{16}}=\frac{\sqrt{16-b^2}}{4}\) So, the focus will be \(\left( \pm \sqrt{16-b^2}, 0\right)\) Also, the equation of hyperbola,…