For the hyperbola frac{x^2}{cos^2 alpha} - fr | vedclass

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(D) Given the hyperbola equation $\frac{x^2}{\cos^2 \alpha} - \frac{y^2}{\sin^2 \alpha} = 1$.Here,$a^2 = \cos^2 \alpha$ and $b^2 = \sin^2 \alpha$.The

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Abscissae of foci

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(D) Given the hyperbola equation $\frac{x^2}{\cos^2 \alpha} - \frac{y^2}{\sin^2 \alpha} = 1$.Here,$a^2 = \cos^2 \alpha$ and $b^2 = \sin^2 \alpha$.The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{\sin^2 \alpha}{\cos^2 \alpha}} = \sqrt{1 + 	an^2 \alpha} = \sec \alpha$.The coordinates of the foci are $(\pm ae, 0)$.Calculating the abscissa of the foci: $ae = \sqrt{\cos^2 \alpha} 	imes \sec \alpha = \cos \alpha 	imes \frac{1}{\cos \alpha} = 1$.Since the abscissa of the foci is $1$,which is independent of $\alpha$,it remains constant.

For the hyperbola $\frac{x^2}{\cos^2 \alpha} - \frac{y^2}{\sin^2 \alpha} = 1$,which of the following remains constant if $\alpha$ varies?

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