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: Understanding the Concept:
"Coinciding foci" means they have the same $ae$ value.
We find foci of ellipse first, then use hyperbola eccentricity to find its semi-axes.
Step 2: Key Formula or Approach:
Ellipse: $a_e^2 = 25, b_e^2 = 9$. Focus $c = \sqrt{a_e^2 - b_e^2} = 4$.
Hyperbola: $c = a_h e_h = 4$. Given $e_h = 2$.
Relation $b^2 = a^2(e^2 - 1)$.
Step 3: Detailed Explanation:
$a_h(2) = 4 \implies a_h = 2 \implies a_h^2 = 4$.
$b_h^2 = 4(2^2 - 1) = 4(3) = 12$.
Equation: $\frac{x^2}{4} - \frac{y^2}{12} = 1$.
Step 4: Final Answer:
The equation is $\frac{x^2}{4} - \frac{y^2}{12} = 1$.