If the ellipse 4x^2 + 9y^2 = 36 is confocal w | vedclass

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(D) Given ellipse: $4x^2 + 9y^2 = 36$,which can be written as $\frac{x^2}{9} + \frac{y^2}{4} = 1$. Here,$a^2 = 9$ and $b^2 = 4$. For the ellipse,$e^2

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$x^2 + y^2 = 5$

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(D) Given ellipse: $4x^2 + 9y^2 = 36$,which can be written as $\frac{x^2}{9} + \frac{y^2}{4} = 1$. Here,$a^2 = 9$ and $b^2 = 4$. For the ellipse,$e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{4}{9} = \frac{5}{9}$. The foci are $(\pm ae, 0) = (\pm 3 	imes \frac{\sqrt{5}}{3}, 0) = (\pm \sqrt{5}, 0)$. Thus,$ae = \sqrt{5}$,so $(ae)^2 = 5$. For the hyperbola,the length of the transverse axis is $2$,so $2a_h = 2 \implies a_h = 1$. Let the hyperbola be $\frac{x^2}{1} - \frac{y^2}{b_h^2} = 1$. Since it is confocal with the ellipse,its foci are also $(\pm \sqrt{5}, 0)$. For a hyperbola,$a_h^2 e_h^2 = a_h^2 + b_h^2$. Here,$a_h^2 e_h^2 = 5$ and $a_h^2 = 1$,so $5 = 1 + b_h^2 \implies b_h^2 = 4$. The hyperbola equation is $x^2 - \frac{y^2}{4} = 1$,or $4x^2 - y^2 = 4$. Adding the ellipse equation $4x^2 + 9y^2 = 36$ and the hyperbola equation $4x^2 - y^2 = 4$: $(4x^2 + 9y^2) + (4x^2 - y^2) = 36 + 4 8x^2 + 8y^2 = 40 8(x^2 + y^2) = 40 x^2 + y^2 = 5$. Thus,the points of intersection lie on the circle $x^2 + y^2 = 5$.

If the ellipse $4x^2 + 9y^2 = 36$ is confocal with a hyperbola whose length of the transverse axis is $2$,then the points of intersection of the ellipse and hyperbola lie on the circle:

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