Question

Let \(x\) be the eccentricity of a hyperbola whose transverse axis is twice its conjugate axis. Let \(y\) be the eccentricity of another hyperbola for which the distance between the foci is 3 times the distance between its directrices. Then \(y^2 - x^2 =\)

• A. \(\frac{23}{4}\)
• B. \(\frac{7}{4}\)
• C. \(\frac{4}{7}\)
• D. \(\frac{16}{23}\)

Hint:

Remember the standard distances for a hyperbola: Distance between foci = \(2ae\), distance between directrices = \(2a/e\).

Answer 1

The Correct Option is B

Step 1: Calculate Eccentricity \(x\) (First Hyperbola):
Given: Transverse axis (\(2a\)) is twice the conjugate axis (\(2b\)). \(2a = 2(2b) \implies a = 2b\). Eccentricity formula: \(x = \sqrt{1 + \frac{b^2}{a^2}}\). Substitute \(a = 2b\): \(x = \sqrt{1 + \frac{b^2}{(2b)^2}} = \sqrt{1 + \frac{1}{4}} = \sqrt{\frac{5}{4}}\). So, \(x^2 = \frac{5}{4}\).
Step 2: Calculate Eccentricity \(y\) (Second Hyperbola):
Given: Distance between foci (\(2ae\)) is 3 times distance between directrices (\(2a/e\)). \(2ae = 3 \times \frac{2a}{e}\). Cancel \(2a\): \(e = \frac{3}{e} \implies e^2 = 3\). So, \(y^2 = 3\).
Step 3: Calculate \(y^2 - x^2\):
\[ y^2 - x^2 = 3 - \frac{5}{4} = \frac{12 - 5}{4} = \frac{7}{4} \]