Let the equations of two ellipses be $E_{1} : \frac{x^{2}}{3}+\frac{y^{2}}{2}=1 and E_{2} : \frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1$, If the product of their eccentricities is $\frac{1}{2}$, then the length of the minor axis of ellipse $E_2$ is:

Question

The area bounded by $ y - 1 = |x| $ and $ y + 1 = |x| $ is: (a) $ \frac{1}{2} $

Answer 1

To solve this problem, we need to find the length of the minor axis of ellipse $E_2$. Let's start by analyzing both ellipses:

1. The equation of ellipse $E_1$ is given by:

$E_1: \frac{x^2}{3} + \frac{y^2}{2} = 1$.

2. The equation of ellipse $E_2$ is:

$E_2: \frac{x^2}{16} + \frac{y^2}{b^2} = 1$.

The eccentricity $e$ of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by:

$e = \sqrt{1 - \frac{b^2}{a^2}}$, where $a > b$.

3. For ellipse $E_1$, $a^2 = 3$ and $b^2 = 2$:

$e_1 = \sqrt{1 - \frac{2}{3}} = \sqrt{\frac{1}{3}}$.

4. For ellipse $E_2$, $a^2 = 16$ and $b^2 = b^2$:

$e_2 = \sqrt{1 - \frac{b^2}{16}}$.

5. We know that the product of their eccentricities is $ \frac{1}{2} $, so:

$e_1 \times e_2 = \frac{1}{2}$.

Substituting the values of $e_1$ and $e_2$:

$\sqrt{\frac{1}{3}} \times \sqrt{1 - \frac{b^2}{16}} = \frac{1}{2}$.

Simplifying, we get:

$\sqrt{\frac{1 - \frac{b^2}{16}}{3}} = \frac{1}{2}$.

Squaring both sides, we have:

$\frac{1 - \frac{b^2}{16}}{3} = \frac{1}{4}$.

Solving this equation:

$1 - \frac{b^2}{16} = \frac{3}{4}$.

$\frac{b^2}{16} = \frac{1}{4}$.

$b^2 = 16 \times \frac{1}{4} = 4$.

Hence, the length of the minor axis of ellipse $E_2$, which is $2b$, is:

$2 \times \sqrt{4} = 4$.

Therefore, the correct answer is $4$.

The Correct Option is C