An ellipse passes through the foci of the hyp | vedclass

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(C) The equation of the hyperbola is $\frac{x^2}{4} - \frac{y^2}{9} = 1$.Its foci are $(\pm \sqrt{4+9}, 0) = (\pm \sqrt{13}, 0)$.The eccentricity of t

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$\left( \frac{\sqrt{13}}{2}, \frac{\sqrt{3}}{2} \right)$

Vedclass · Answer

(C) The equation of the hyperbola is $\frac{x^2}{4} - \frac{y^2}{9} = 1$.Its foci are $(\pm \sqrt{4+9}, 0) = (\pm \sqrt{13}, 0)$.The eccentricity of the hyperbola is $e_h = \frac{\sqrt{13}}{2}$.Let $e_e$ be the eccentricity of the ellipse. Given $e_e 	imes e_h = \frac{1}{2}$,we have $e_e 	imes \frac{\sqrt{13}}{2} = \frac{1}{2}$,so $e_e = \frac{1}{\sqrt{13}}$.The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Since the ellipse passes through the foci of the hyperbola $(\pm \sqrt{13}, 0)$,we have $a^2 = 13$.Using $b^2 = a^2(1 - e_e^2)$,we get $b^2 = 13(1 - \frac{1}{13}) = 13 - 1 = 12$.The equation of the ellipse is $\frac{x^2}{13} + \frac{y^2}{12} = 1$.Checking the points:For $A$: $\frac{13/2}{13} + \frac{6}{12} = \frac{1}{2} + \frac{1}{2} = 1$ (Lies on ellipse).For $B$: $\frac{39/4}{13} + \frac{3}{12} = \frac{3}{4} + \frac{1}{4} = 1$ (Lies on ellipse).For $C$: $\frac{13/4}{13} + \frac{3/4}{12} = \frac{1}{4} + \frac{1}{16} = \frac{5}{16} 
eq 1$ (Does not lie on ellipse).For $D$: $\frac{13}{13} + 0 = 1$ (Lies on ellipse).Thus,the point in option $C$ does not lie on the ellipse.

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