Question

The ellipse $x^2+4y^2=4$ is inscribed in a rectangle aligned with the coordinate axes, which in turn in 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

Answer: (B)

$x^2+4y^2=4\Rightarrow \frac{x^2}{4}+\frac{y^2}{1}=1\Rightarrow a=2,b=1\Rightarrow P=\left(2,1\right)$
Required Ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\Rightarrow \frac{x^2}{4^2}+\frac{y^2}{b^2}=1$
$\left(2,1\right)$ lies on it
$\Rightarrow \frac{4}{16}+\frac{1}{b^2}=1\Rightarrow \frac{1}{b^2}=1-\frac{1}{4}=\frac{3}{4}\Rightarrow b^2=\frac{4}{3}$
$\therefore \frac{x^2}{16}+\frac{y^2}{\left(\frac{4}{3}\right)}=1\Rightarrow \frac{x^2}{16}+\frac{3y^2}{4}=1\Rightarrow x^2+12y^2=16$