Question

The foci of a hyperbola coincide with the foci of the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$. The equation of the hyperbola with eccentricity 2 is

• A. $\frac{x^2}{12} - \frac{y^2}{4} = 1$
• B. $\frac{x^2}{4} - \frac{y^2}{12} = 1$
• C. $\frac{x^2}{12} - \frac{y^2}{16} = 1$
• D. $\frac{x^2}{16} - \frac{y^2}{12} = 1$

Hint:

Coinciding foci means the value of $ae$ is identical for both curves.

Answer 1

The Correct Option is B

Step 1: Find Foci of Ellipse
For ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$, $a^2=25, b^2=9$.
$e_e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}$.
Foci $(\pm ae_e, 0) = (\pm 5 \cdot \frac{4}{5}, 0) = (\pm 4, 0)$.
Step 2: Hyperbola Parameters
Foci of hyperbola are $(\pm ae_h, 0) = (\pm 4, 0)$.
Given $e_h = 2$, so $a \cdot 2 = 4 \implies a = 2 \implies a^2 = 4$.
Step 3: Find $b^2$ for Hyperbola
$b^2 = a^2(e_h^2 - 1) = 4(2^2 - 1) = 4(3) = 12$.
Equation: $\frac{x^2}{4} - \frac{y^2}{12} = 1$.
Final Answer: (B)