Q. The ellipse $x^{2} + 4y^{2} = 4$ is inscribed in a rectangle aligned with the coordinate axes, which is turn in inscribed in another ellipse that passes through the point (4, 0). Then, the equation of the ellipse is
A) $x^{2} + 12y^{2} = 16$
B) $4x^{2} + 48y^{2} = 48$
C) ${4x}^{2} + 64y^{2} = 48$
D) $x^{2} + 16y^{2} = 16$
Explanation
(a)
Let the equation of the required ellipse be $\frac{x^{2}}{16} + \frac{y^{2}}{b^{2}} = 1$
But the ellipse passes through the point $(2,1)$
[image:11_image63.emf]
$\Rightarrow \ \ \ \frac{1}{4} + \frac{1}{b^{2}} = 1$
$\Rightarrow \ \ \frac{1}{b^{2}} = \frac{3}{4}\ \Rightarrow \ b^{2} = \frac{4}{3}$
Hence, equation is
$\frac{x^{2}}{16} + \frac{3y^{2}}{4} = 1$
$\Rightarrow \ x^{2} + 12y^{2} = 16$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.