Question

The ellipse $x^2+4y^2=4$ is inscribed in a rectangle aligned with the coordinate axes, which is turn in inscribed in another ellipse that passes through the point $\left(4,0\right)$. Then, the equation of the ellipse is

• A. $x^2+12y^2=16$
• B. $4x^2+48y^2=48$
• C. $4x^2+64y^2=48$
• D. $x^2+16y^2=16$

Answer 1

The correct option is A

$x^2+12y^2=16$

Explanation for the correct option:

Given,

The equation of the ellipse is:

$$\begin{gathered} x^2+4y^2=4 \\ \Rightarrow \frac{x^2}{4}+\frac{y^2}{1}=1 \end{gathered}$$

Let equation of the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Since this ellipse passes through$\left(4,0\right)$

$$\begin{gathered} \frac{4^2}{a^2}+\frac{0}{b^2}=1 \\ \Rightarrow \frac{16}{a^2}=1 \\ \Rightarrow a^2=16 \end{gathered}$$

since the ellipse $x^2+4y^2=4$ is inscribed in a rectangle passing through $\left(2,1\right)$ the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$also passes through $\left(2,1\right)$therefore

$$\begin{gathered} \frac{2^2}{16}+\frac{1^2}{b^2}=1 \\ \Rightarrow \frac{1}{4}+\frac{1}{b^2}=1 \\ \Rightarrow \frac{1}{b^2}=\frac{3}{4} \\ \Rightarrow b^2=\frac{4}{3} \end{gathered}$$

So the equation of the required ellipse is $\frac{x^2}{16}+\frac{3y^2}{4}=1$ That is $x^2+12y^2=16$

Hence, option (A) is the correct answer