A hyperbola,having the transverse axis of len | vedclass

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Question

(A) The given ellipse is $\frac{x^2}{4} + \frac{y^2}{3} = 1$. Comparing with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 4$ and $b^2 = 3$.

Vedclass · Accepted Answer

$x^2 \operatorname{cosec}^2 	heta - y^2 \sec^2 	heta = 1$

Vedclass · Answer

(A) The given ellipse is $\frac{x^2}{4} + \frac{y^2}{3} = 1$. Comparing with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 4$ and $b^2 = 3$. The eccentricity $e$ of the ellipse is $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{3}{4}} = \frac{1}{2}$. The foci are $(\pm ae, 0) = (\pm 2 	imes \frac{1}{2}, 0) = (\pm 1, 0)$. For the hyperbola,the transverse axis length is $2a_1 = 2 \sin 	heta$,so $a_1 = \sin 	heta$. Since the hyperbola is confocal with the ellipse,its foci are $(\pm a_1 e_1, 0) = (\pm 1, 0)$,so $a_1 e_1 = 1$. Thus,$e_1 = \frac{1}{\sin 	heta} = \operatorname{cosec} 	heta$. For a hyperbola,$b_1^2 = a_1^2 (e_1^2 - 1) = \sin^2 	heta (\operatorname{cosec}^2 	heta - 1) = \sin^2 	heta \cot^2 	heta = \cos^2 	heta$. The equation of the hyperbola is $\frac{x^2}{a_1^2} - \frac{y^2}{b_1^2} = 1$,which is $\frac{x^2}{\sin^2 	heta} - \frac{y^2}{\cos^2 	heta} = 1$,or $x^2 \operatorname{cosec}^2 	heta - y^2 \sec^2 	heta = 1$.

$A$ hyperbola,having the transverse axis of length $2 \sin \theta$,is confocal with the ellipse $3 x^2 + 4 y^2 = 12$. Then its equation is

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