Question

Let \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where \( a>b \), be an ellipse whose eccentricity is \( \frac{1}{\sqrt{2}} \) and the length of the latus rectum is \( \sqrt{14} \). Then the square of the eccentricity of \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) is:

• A. 3
• B. \(\frac{7}{2}\)
• C. \(\frac{3}{2}\)
• D. \(\frac{5}{2}\)

Answer 1

The Correct Option is C

For an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) with \( a>b \), given its eccentricity \( e_1 = \frac{1}{\sqrt{2}} \) and latus rectum length \( \sqrt{14} \), determine the square of the eccentricity of the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).

Concept Used:

For an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) where \( a>b \), the eccentricity \( e_1 \) is related by \( b^2 = a^2(1 - e_1^2) \), and the latus rectum length is \( \frac{2b^2}{a} \).
For a hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the eccentricity \( e_2 \) satisfies \( b^2 = a^2(e_2^2 - 1) \).

Step-by-Step Solution:

Step 1: Relate \( a \) and \( b \) using the ellipse's eccentricity.

\[ e_1 = \frac{1}{\sqrt{2}} \quad \Rightarrow \quad b^2 = a^2(1 - e_1^2) = a^2\left(1 - \frac12\right) = \frac{a^2}{2} \]

Step 2: Use the latus rectum length of the ellipse.

\[ \text{Latus rectum} = \frac{2b^2}{a} = \sqrt{14} \] \[ \frac{2 \cdot \frac{a^2}{2}}{a} = \sqrt{14} \quad \Rightarrow \quad \frac{a^2}{a} = \sqrt{14} \quad \Rightarrow \quad a = \sqrt{14} \]

Step 3: Calculate \( b^2 \).

\[ b^2 = \frac{a^2}{2} = \frac{14}{2} = 7 \]

Step 4: Determine the eccentricity \( e_2 \) of the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).

For this hyperbola, \( b^2 = a^2(e_2^2 - 1) \).

\[ 7 = 14(e_2^2 - 1) \quad \Rightarrow \quad e_2^2 - 1 = \frac12 \quad \Rightarrow \quad e_2^2 = \frac12 + 1 = \frac32 \]

Step 5: The square of the eccentricity is \( e_2^2 \).

Thus, the square of the eccentricity of the hyperbola is \( \mathbf{\frac{3}{2}} \).