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The ellipse E1:(x^2)/9+(y^2)/4=1 is insc...

The ellipse `E_1:(x^2)/9+(y^2)/4=1` is inscribed in a rectangle `R` whose sides are parallel to the coordinate axes. Another ellipse `E_2` passing through the point (0, 4) circumscribes the rectangle `Rdot` The eccentricity of the ellipse `E_2` is `(sqrt(2))/2` (b) `(sqrt(3))/2` (c) `1/2` (d) `3/4`

A

`sqrt(2)//2`

B

`sqrt(3)//2`

C

`1//2`

D

`3//4`

Text Solution

Verified by Experts

Let the ellipse be
`(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`
as it passing through (0,4) and (3,2). So,
`b^(2)=16 and (9)/(a^(2))+(4)+(16)=1`
or `a^(2)=12` ltbr So, `12=16(1-e^(2))`
or `e=(1)/(2)`
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