Question

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

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

Answer 1

The correct option is B $x^2+12y^2=16$

Given ellipse $x^2+4y^2=4$

$\frac{x^2}{4}+\frac{y^2}{1}=1\dots \left(1\right)$

Given that second ellipse is passing through $(4,0)$

Substituting $(4,0)$ in $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\frac{4^2}{a^2}+\frac{0^2}{b^2}=1\dots \left(1\right)$

so the semi-major axis of the ellipse is $4$.

If length of semi-minor axis is $b$ then $\frac{x^2}{16}+\frac{y^2}{b^2}=1$

It passes through $(2,1)$ So $\frac{4}{16}+\frac{1}{b^2}=1$

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

$\Rightarrow b^2=\frac{4}{3}$

Substituting $a^2=16$ and $b^2=\frac{4}{3}in\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

We get,

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