Question

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

$x^2+4y^2=4$
$\Rightarrow \frac{x^2}{4}+\frac{y^2}{1}=1$
So, $a=2,b=1$.
Thus $P$ is $(2,1)$.
The required ellipse is
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$\Rightarrow \frac{x^2}{4^2}+\frac{y^2}{b^2}=1$
The point $(2,1)$ lies on it. So
$\frac{4}{16}+\frac{1}{b^2}=1$
$\Rightarrow \frac{1}{b^2}=1-\frac{1}{4}=\frac{3}{4}$
$\Rightarrow b^2=\frac{4}{3}$
$\therefore \frac{x^2}{16}+\frac{y^2}{(\frac{4}{3})}=1$
$\Rightarrow \frac{x^2}{16}+\frac{2y^2}{4}=1$
$\Rightarrow x^2+12y^2=16$