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
$\frac{4}{16} + \frac{1}{b^2} = 1$
$\Rightarrow \frac{1}{b^2} = 1 - \frac{1}{4} = \frac{3}{4}$
$\Rightarrow b^2 = \frac{4}{3}$
$\therefore \frac{x^2}{16} + \frac{y^2}{(\frac{4}{3})} = 1$
$\Rightarrow \frac{x^2}{16} + \frac{2y^2}{4} = 1$
$\Rightarrow x^2 + 12y^2 = 16$