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
- A \(5\)
- B \(7\)
- C \(9\)
- D \(1\)
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,…
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