Question

Let the foci of a hyperbola $H$ coincide with the foci of the ellipse $E : \frac{(x - 1)^2}{100} + \frac{(y - 1)^2}{75} = 1$ and the eccentricity of the hyperbola $H$ be the reciprocal of the eccentricity of the ellipse $E$. If the length of the transverse axis of $H$ is $\alpha$ and the length of its conjugate axis is $\beta$, then $3\alpha^2 + 2\beta^2$ is equal to:

• A. 242
• B. 225
• C. 237
• D. 205

Answer 1

Answer: (B)

225

Answer 2

We are given an ellipse with the following properties:

$e_1 = \sqrt{1 - \frac{75}{100}} = \sqrt{\frac{5}{10}} = \frac{1}{2}$

$e_2 = 2$

The foci are $F_1(6, 1)$ and $F_2(-4, 1)$.

We proceed with the following steps:

$2ae_2 = 10 \implies a = \frac{5}{2}$

$\alpha = 5$

$4 = 1 + \frac{b^2}{a^2} \implies b^2 = 3a^2 \implies b = \sqrt{3} \times \frac{5}{2}$

Thus,

$\beta = 5\sqrt{3}$

$3\alpha^2 + 2\beta^2 = 3 \times 25 + 2 \times 25 \times 3 = 225$

Thus, the area is 225.