Question

The ellipse$E1:\frac{x^2}{9}+\frac{y^2}{4}=1$ is inscribed in a rectangle $R$ whose sides are parallel to the coordinate axes. Another ellipse $E_2$ passing through the point $(0,4)$ circumscribes the rectangle $R$. The eccentricity of the ellipse $E_2$ is

• A. $\frac{\sqrt{2}}{2}$
• B. $\frac{\sqrt{3}}{2}$
• C. $\frac{1}{2}$
• D. $\frac{3}{4}$

Answer 1

The correct option is C

$\frac{1}{2}$

Explanation for the correct option:

Find the value of eccentricity:

Given,

$E1:\frac{x^2}{9}+\frac{y^2}{4}=1\to \left(i\right)$

We know that the general equation of the ellipse whose center is origin is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ ,

Then the ellipse touches the x-axis at $\left(\pm a,0\right)$ and y-axis at $\left(0,\pm b\right)$

Then the ellipse $E_1$ touches the x-axis at $\left(\pm 3,0\right)$ and y-axis at $\left(0,\pm 2\right)$.

Given,

The ellipse is inscribed in a rectangle $R$ whose sides are parallel to the coordinate axes

Therefore the vertices of the rectangle are$\left(\pm 3,\pm 2\right)$.

Another ellipse $E_2$ circumscribes the rectangle $R$.and is passing through the point $(0,4)$

Which means $E_2$ passing through the point $\left(\pm 3,\pm 2\right)$ and $(0,4)$

Substitute the point $(0,4)$in the general form of an ellipse.

Then

$$\begin{gathered} \frac{0}{a^2}+\frac{4^2}{b^2}=1 \\ \Rightarrow b^2=4^2 \end{gathered}$$

Substitute $b^2=4^2$ and the point $\left(\pm 3,\pm 2\right)$ in the general equation.

We get,

$$\begin{gathered} \frac{3^2}{a^2}+\frac{2^2}{4^2}=1 \\ \Rightarrow \frac{9}{a^2}=1-\frac{1}{4} \\ \Rightarrow \frac{9}{a^2}=\frac{3}{4} \\ \Rightarrow a^2=\frac{9\times 4}{3} \\ \Rightarrow a^2=12 \end{gathered}$$

Find eccentricity:

We know that

$$\begin{gathered} a^2=b^2\left(1-e^2\right) \\ \Rightarrow 12=16\left(1-e^2\right) \\ \Rightarrow \frac{12}{16}=1-e^2 \\ \Rightarrow e^2=1-\frac{3}{4} \\ \Rightarrow e^2=\frac{1}{4} \\ \Rightarrow e=\frac{1}{2} \end{gathered}$$

Hence, option (C) is the correct answer.