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Question: For hyperbola \(\frac{x^{2}}{\cos^{2}\alpha} - \frac{y^{2}}{\sin^{2}\alpha} = 1\) which of the follo...

For hyperbola x2cos⁡2α−y2sin⁡2α=1\frac{x^{2}}{\cos^{2}\alpha} - \frac{y^{2}}{\sin^{2}\alpha} = 1 which of the following remains constant with change in 'α':

A

Abscissa of vertices

B

Abscissa of foci

C

Eccentricity

D

Directrix

Answer

Abscissa of foci

Explanation

Solution

e2 = 1 + b2a2\frac{b^{2}}{a^{2}} ⇒ e2 = 1 + sin⁡2αcos⁡2α=sec⁡2α\frac{\sin^{2}\alpha}{\cos^{2}\alpha} = \sec^{2}\alpha

Now foci (± ae, 0) ⇒ (±1, 0).