ExamBro
ExamBro
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

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

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…