The ellipse $x^2+4 y^2=4$ 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

  $x^2+16 y^2=16$
• Correct answer
  B

  $x^2+12 y^2=16$
• C

  $4 x^2+48 y^2=48$
• D

  $4 \mathrm{x}^2+64 \mathrm{y}^2=48$

Solution

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

Hence, the correct answer is option: 'B'.