For the hyperbola x^2 sec^2 theta - y^2 csc^2 | vedclass

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(A) The given equation is $x^2 \sec^2 	heta - y^2 \csc^2 	heta = 1$,which can be written as $\frac{x^2}{\cos^2 	heta} - \frac{y^2}{\sin^2 	heta} =

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(A) The given equation is $x^2 \sec^2 	heta - y^2 \csc^2 	heta = 1$,which can be written as $\frac{x^2}{\cos^2 	heta} - \frac{y^2}{\sin^2 	heta} = 1$.Comparing this with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we get $a^2 = \cos^2 	heta$ and $b^2 = \sin^2 	heta$.The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{\sin^2 	heta}{\cos^2 	heta}} = \sqrt{1 + 	an^2 	heta} = \sqrt{\sec^2 	heta} = \sec 	heta$.The coordinates of the foci are $(\pm ae, 0)$.$ae = \sqrt{a^2} \cdot e = \sqrt{\cos^2 	heta} \cdot \sec 	heta = \cos 	heta \cdot \frac{1}{\cos 	heta} = 1$.Since the distance of the foci from the center is $ae = 1$,which is independent of $	heta$,the foci remain constant.

For the hyperbola $x^2 \sec^2 \theta - y^2 \csc^2 \theta = 1$,which of the following remains constant when $\theta$ varies?

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