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The ellipse ${x^2} + 4{y^2} = 4$ is inscribed in a rectangle aligned with the coordinate axes, which in trun is inscribed in another ellipse that passes through the point $(4,0) $ . Then the equation of the ellipse is :
$\;{x^2} + 12{y^2} = 16$
$\;4{x^2} + 48{y^2} = 48$
$\;4{x^2} + 64{y^2} = 48$
$\;{x^2} + 16{y^2} = 16$
Solution
$x^{2}+4 y^{2}=4;$
$\frac{x^{2}}{4}+\frac{y^{2}}{1}=1;$
So, $a^{2}=4$ and $b^{2}=1;$
So, $a=2$ and $b=1;$
So $P=(2,1);$
Equation of ellipse is,
$\frac{x^{2}}{4^{2}}+\frac{y^{2}}{b^{2}}=1 ; \ldots .(1)$
It passes throgh $(2,1),$
$\frac{2^{2}}{4^{2}}+\frac{1}{b^{2}}=1;$
$\frac{1}{b^{2}}=1-\frac{1}{4}=\frac{3}{4};$
$b^{2}=\frac{4}{3};$
Substituting this in $( 1),$
$\frac{x^{2}}{4^{2}}+\frac{y^{2}}{\left(\frac{4}{3}\right)}=1$
$\frac{x^{2}}{16}+\frac{3 y^{2}}{4}=1$
$x^{2}+12 y^{2}=16$
