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

Answer: (C)

$\frac{3}{2}$

Answer 2

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 $\frac{3}{2}$.