Question

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

• A. 4
• B. 9
• C. 14
• D. 7

Hint:

For conic sections, the foci are determined by \( c = \sqrt{a^2 \pm b^2} \), and this relationship can help solve for unknown parameters when the foci coincide.

Answer 1

The Correct Option is D

Step 1: Analyze the foci of the ellipse and hyperbola.
For the ellipse, the foci are given by \( c = \sqrt{a^2 - b^2} \), and for the hyperbola, the foci are given by \( c = \sqrt{a^2 + b^2} \). Set the two expressions for \( c \) equal to each other because the foci coincide.
Step 2: Solve for \( b^2 \).
After solving, we find that \( b^2 = 7 \).
Step 3: Conclusion.
The correct answer is (D) 7.