If a hyperbola has a transverse axis of lengt | vedclass

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(A) The given ellipse is $3x^2 + 4y^2 = 12$,which can be written as $\frac{x^2}{4} + \frac{y^2}{3} = 1$.Here,$a^2 = 4$ and $b^2 = 3$.The eccentricity

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$x^2 \csc^2 	heta - y^2 \sec^2 	heta = 1$

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(A) The given ellipse is $3x^2 + 4y^2 = 12$,which can be written as $\frac{x^2}{4} + \frac{y^2}{3} = 1$.Here,$a^2 = 4$ 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 length of the transverse axis is $2a_h = 2 \sin 	heta$,so $a_h = \sin 	heta$.Since the hyperbola is confocal with the ellipse,its foci are $(\pm 1, 0)$,so $a_h e_h = 1$.Thus,$e_h = \frac{1}{\sin 	heta} = \csc 	heta$.For the hyperbola,$b_h^2 = a_h^2(e_h^2 - 1) = \sin^2 	heta (\csc^2 	heta - 1) = \sin^2 	heta \cot^2 	heta = \cos^2 	heta$.The equation of the hyperbola is $\frac{x^2}{a_h^2} - \frac{y^2}{b_h^2} = 1$,which is $\frac{x^2}{\sin^2 	heta} - \frac{y^2}{\cos^2 	heta} = 1$.This simplifies to $x^2 \csc^2 	heta - y^2 \sec^2 	heta = 1$.

If a hyperbola has a transverse axis of length $2 \sin \theta$ and is confocal with the ellipse $3x^2 + 4y^2 = 12$,then its equation is:

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