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the ellipse x 2 4y 2 4 is inscribed in a rectangle
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
AIEEE - 2009
AIEEE
Updated On:
Aug 15, 2022
$x^2 +16y^2 = 16$
$x^2 +12y^2 = 16$
$4x^2 + 48y^2 = 48$
$4x^2 + 64y^2 = 48$
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The Correct Option is
B
Solution and Explanation
$x^{2}+4y^{2}=4 \Rightarrow \frac{x^{2}}{4}+\frac{y^{2}}{1}=1 \Rightarrow a=2, b=1 \Rightarrow P=\left(2, 1\right)$ 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 $\Rightarrow \frac{4}{16}+\frac{1}{b^{2}}=1 \Rightarrow \frac{1}{b^{2}}=1-\frac{1}{4}=\frac{3}{4} \Rightarrow b^{2}=\frac{4}{3}$ $\therefore \frac{x^{2}}{16}+\frac{y^{2}}{\left(\frac{4}{3}\right)}=1 \Rightarrow \frac{x^{2}}{16}+\frac{3y^{2}}{4}=1 \Rightarrow x^{2}+12y^{2}=16$
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