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

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(B) The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,where $a^2 = \cos^2 \alpha$ and $b^2 = \sin^2 \alpha$.For a hyperbola,the

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Abscissae of foci

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(B) The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,where $a^2 = \cos^2 \alpha$ and $b^2 = \sin^2 \alpha$.For a hyperbola,the foci are located at $(\pm ae, 0)$,where $e$ is the eccentricity.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 abscissa of the foci is $\pm ae = \pm \sqrt{\cos^2 \alpha} \cdot \sec \alpha = \pm \cos \alpha \cdot \frac{1}{\cos \alpha} = \pm 1$.Since the value $\pm 1$ is independent of $\alpha$,the abscissae of the foci remain constant.

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

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