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

To solve this problem, we need to understand the relationships and characteristics of the ellipse and hyperbola given.

We start with the equation of the ellipse \(E: \frac{(x - 1)^2}{100} + \frac{(y - 1)^2}{75} = 1\).

1. The standard form of the ellipse equation is \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), where \(a^2\) is the denominator of the term with the longer axis.
2. Here, \(h = 1\) and \(k = 1\), hence the center is \((1, 1)\).
3. Given \(a^2 = 100\) and \(b^2 = 75\), we identify that \(a > b\), so the major axis is along the x-axis with \(a = 10\) and \(b = \sqrt{75}\).
4. The eccentricity \(e\) of the ellipse is given by \(e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{75}{100}} = \sqrt{0.25} = 0.5\).
5. The foci of the ellipse are located at \((h \pm ae, k)\), so the foci are \((1 \pm 5, 1)\) => \((-4, 1)\) and \((6, 1)\).

Next, we move on to the hyperbola \(H\) which has the same foci as ellipse \(E\). The eccentricity of hyperbola \(H\) is the reciprocal of the eccentricity of ellipse \(E\).

1. The eccentricity \(e_H\) for hyperbola \(H\) is given as the reciprocal of the ellipse's eccentricity. Thus, \(e_H = \frac{1}{0.5} = 2\).
2. The standard form of a hyperbola is \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), where \(c = ae_H\).
3. From the foci, we know \(c = 5\), therefore \(5 = 2a\) => \(a = 2.5\).
4. Given the eccentricity formula \(e_H = \frac{c}{a}\), find \(b\) using \(c^2 = a^2 + b^2\). So, \(25 = 2.5^2 + b^2 => b^2 = 25 - 6.25 = 18.75\).

We find the lengths of the transverse and conjugate axes from the values of \(a\) and \(b\):

The length of the transverse axis \(H\) is \(\alpha = 2a = 2 \times 2.5 = 5\).

The length of the conjugate axis \(H\) is \(\beta = 2b = 2 \times \sqrt{18.75} = 2 \times \sqrt{75/4} = 2 \times \frac{\sqrt{75}}{2} = \sqrt{75}\approx 8.66\).

Finally, we compute \(3\alpha^2 + 2\beta^2\):

\(\alpha^2 = 25\) and \(\beta^2 = 75\).

\(3\alpha^2 + 2\beta^2 = 3 \times 25 + 2 \times 75 = 75 + 150 = 225\).

Therefore, the final answer is 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.