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

The Correct Option is B

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.