Ellipse {x^2} + 4{y^2} = 4 is inscribed in a rectangle aligned with coordinate axes, which in trun i

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10-2. Parabola, Ellipse, Hyperbola

hard

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 :

$\;{x^2} + 12{y^2} = 16$

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

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

$\;{x^2} + 16{y^2} = 16$

(AIEEE-2009)

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 $( 1),$

$\frac{x^{2}}{4^{2}}+\frac{y^{2}}{\left(\frac{4}{3}\right)}=1$

$\frac{x^{2}}{16}+\frac{3 y^{2}}{4}=1$

$x^{2}+12 y^{2}=16$

Standard 11

Mathematics

Which one of the following is the common tangent to the ellipses, $\frac{{{x^2}}}{{{a^2} + {b^2}}} + \frac{{{y^2}}}{{{b^2}}}$ $=1\&$ $ \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{a^2} + {b^2}}}$ $=1$

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(JEE MAIN-2017)

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 :

Solution

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