KVPY Sample Paper KVPY Stream-SX Model Paper-9

For hyperbola \[\frac{{{x}^{2}}}{{{\cos }^{2}}\alpha }-\frac{{{y}^{2}}}{{{\sin }^{2}}\alpha }=1\]which of the following remains constant with change in \['\alpha '\]

A) abscissae of vertices

B) abscissae of foci

C) Eccentricity

D) directrix

Correct Answer: B

Solution :

The given \[e{{q}^{n}}\]of hyperbola is \[\frac{{{x}^{2}}}{{{\cos }^{2}}\alpha }-\frac{{{y}^{2}}}{{{\sin }^{2}}\alpha }=1\]

\[\Rightarrow a=\cos \alpha ,b=\sin \alpha \]

\[\Rightarrow e=\sqrt{1+\frac{{{b}^{2}}}{{{a}^{2}}}=\sqrt{1+{{\tan }^{2}}\alpha }}=\sec \alpha \]

\[\Rightarrow ae=1\]

\[\therefore foci(\pm 1,0)\]

\[\therefore \]foci remain constant with respect to\[\alpha \].