question_answer

The ellipse \[{{x}^{2}}+4{{y}^{2}}=4\]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) \[{{x}^{2}}+16{{y}^{2}}=16\]

B) \[{{x}^{2}}+12{{y}^{2}}=16\]

C) \[4{{x}^{2}}+48{{y}^{2}}=48\]

D) \[4{{x}^{2}}+64{{y}^{2}}=48\]

Correct Answer: B

Solution :

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