COMEDK · Maths · 14. Ellipse
If 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 \(1\)
- B \(7\)
- C \(5\)
- D \(9\)
Answer & Solution
Correct Answer
(B) \(7\)
Step-by-step Solution
Detailed explanation
The equation of the ellipse is \(\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1\) and equation of the hyperbola is \(\frac{x^{2}}{144}-\frac{y^{2}}{81}=\frac{1}{25}\) or \(\frac{x^{2}}{\left(\frac{12}{5}\right)^{2}}-\frac{y}{\left(\frac{9}{5}\right)^{2}}=1\) \(\therefore\) Eccentricity…
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