Question

The ellipse $x^2+4y^2=4$ is inscribed in a rectangle touches its side and aligned with the coordinates axes, which is turn in inscribed in another ellipse which passes through that passes through the point $(4,0).$ Then , the equation of 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 $x^2+12y^2=16$

The equation of the given ellipse is
$\frac{x^2}{4}+\frac{y^2}{1}=1$
The required rectangle must be $x=\pm 2$ and $y=\pm 1$
The second ellipse passes through $(4,0)$
$\Rightarrow a=4$
Now,
Equation of this ellipse can be $\frac{x^2}{16}+\frac{y^2}{b^2}=1\cdots \left(1\right)$
Coordinates of vertices of rectangle will be $(\pm 2,\pm 1)$
$\therefore (2,1)$ will be lying on $E_2$
From equation $\left(1\right)$
$\Rightarrow \frac{4}{16}+\frac{1}{b^2}=1$
$\frac{1}{b^2}=1-\frac{1}{4}=\frac{3}{4}$
$b^2=\frac{4}{3}$
$\therefore$ The equation of ellipse is
$\frac{x^2}{16}+\frac{3y^2}{4}=1$
$x^2+12y^2=16$