COMEDK · Maths · 15. Hyperbola
If the foci of the ellipse \(\dfrac{x^2}{16}+\dfrac{y^2}{b^2}=1\) and the foci of the hyperbola \(\dfrac{x^2}{144}-\dfrac{y^2}{81}=\dfrac{1}{25}\) coincide, then the value of \(b^2\) is
- A \(1\)
- B \(5\)
- C \(7\)
- D \(9\)
Answer & Solution
Correct Answer
(C) \(7\)
Step-by-step Solution
Detailed explanation
The equation of the ellipse is \(\dfrac{x^2}{16} + \dfrac{y^2}{b^2} = 1\). For this ellipse, \(a^2 = 16\). The foci are at \((\pm ae, 0)\), where \(e = \sqrt{1 - \dfrac{b^2}{16}}\). Thus, \(ae = \sqrt{16 - b^2}\). The foci are at \((\pm \sqrt{16 - b^2}, 0)\). The equation of the…
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