Question

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

Answer 1

Step 1. Finding the equation of the ellipse circumscribing the rectangle

The given ellipse equation is is $\frac{x^2}{4}+\frac{y^2}{1}=1.$

Another ellipse that passes through the point $(4,0)$

The corner of the rectangle in the $1st$quadrant, is $(2,1)$.

The ellipse circumscribing the rectangle passes through the point $(4,0)$.

We know that the equation of the ellipse is given by.

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

So, the equation of circumscribing ellipse is $\frac{x^2}{16}+\frac{y^2}{b^2}=1$

Step 2. Find the value of $\frac{1}{b^2}$

$A(2,1)$ lies on the ellipse given above.

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

$$\begin{gathered} \Rightarrow \frac{x^2}{16}+\frac{3}{4}{y}^2=1 \\ \Rightarrow x^2+12y^2=16 \end{gathered}$$

Hence, the required equation is $x^2+12y^2=16$