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)

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

Answer 2

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