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

We are given:

• eccentricity \( e = \frac{1}{\sqrt{2}} \)
• Length of the latus rectum \( L = \sqrt{14} \)

We know the formula for the latus rectum of an ellipse is:

\[ L = \frac{2b^2}{a} \]

From the problem, we are given \( L = \sqrt{14} \), so:

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

Thus,

\[ 2b^2 = a \sqrt{14} \tag{1} \]

Next, use the relationship for the eccentricity \( e \) of an ellipse:

\[ e = \sqrt{1 - \frac{b^2}{a^2}} = \frac{1}{\sqrt{2}} \]

Squaring both sides:

\[ e^2 = 1 - \frac{b^2}{a^2} = \frac{1}{2} \]

This implies:

\[ \frac{b^2}{a^2} = \frac{1}{2} \]

Substitute this into equation (1):

\[ 2 \left(\frac{a^2}{2}\right) = a \sqrt{14} \] \[ a^2 = a \sqrt{14} \]

Thus:

\[ a = \sqrt{14} \]               1

Substitute \( a = \sqrt{14} \) into \( b^2 = \frac{a^2}{2} \):

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

Now, compute the square of the eccentricity:

\[ e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{7}{14} = \frac{7}{14} = \frac{1}{2} \]

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