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 resolve this problem, an understanding of the given ellipse and hyperbola's properties and relationships is required.

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

1. The standard ellipse equation form is \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), where \(a^2\) is the denominator corresponding to the longer axis.
2. The center of the ellipse is determined by \(h = 1\) and \(k = 1\), resulting in a center at \((1, 1)\).
3. With \(a^2 = 100\) and \(b^2 = 75\), it is evident that \(a > b\). This indicates the major axis lies along the x-axis, with semi-axis lengths \(a = 10\) and \(b = \sqrt{75}\).
4. The eccentricity \(e\) of the ellipse is calculated as \(e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{75}{100}} = \sqrt{0.25} = 0.5\).
5. The foci of the ellipse are positioned at \((h \pm ae, k)\), which yields foci at \((1 \pm 5, 1)\) => \((-4, 1)\) and \((6, 1)\).

Subsequently, we analyze the hyperbola \(H\), which shares the same foci as ellipse \(E\). The eccentricity of hyperbola \(H\) is the inverse of the ellipse's eccentricity.

1. The eccentricity \(e_H\) of hyperbola \(H\) is the reciprocal of the ellipse's eccentricity, so \(e_H = \frac{1}{0.5} = 2\).
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 deduce \(c = 5\). Consequently, using \(c = ae_H\), we have \(\Rightarrow 5 = 2a\) => \(\Rightarrow a = 2.5\).
4. Using the eccentricity formula \(e_H = \frac{c}{a}\) and the relation \(c^2 = a^2 + b^2\), we find \(\Rightarrow b^2 = c^2 - a^2\). Substituting the values, we get \(\Rightarrow b^2 = 25 - 2.5^2 = 25 - 6.25 = 18.75\).

The lengths of the transverse and conjugate axes are determined from the calculated values of \(a\) and \(b\):

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

The length of the conjugate axis \(\beta\) for hyperbola \(H\) 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 \(3\alpha^2 + 2\beta^2\):

We have \(\alpha^2 = 5^2 = 25\) and \(\beta^2 = (\sqrt{75})^2 = 75\).

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

The final result is 225.