Question

For hyperbola $\left(\frac{x^2}{{\cos }^2\alpha }\right)-\left(\frac{y^2}{{\sin }^2\alpha }\right)=1$ which of the following remains constant with the change in $‘\alpha ’$

• A. abscissae of vertices
• B. abscissae of foci
• C. eccentricity
• D. directrix

Answer 1

The correct option is B

abscissae of foci

Explanation for the correct option:

Finding which of The following remains constant with the change in $‘\alpha ’$

Given that $\left(\frac{x^2}{{\cos }^2\alpha }\right)-\left(\frac{y^2}{{\sin }^2\alpha }\right)=1$

On comparing with $\left(\frac{x^2}{a^2}\right)-\left(\frac{y^2}{b^2}\right)=1$

$a^2={\cos }^2\alpha \&b^2={\sin }^2\alpha .$

$$\begin{gathered} {\sin }^2\alpha +{\cos }^2\alpha =a^2+b^2 \\ e=\frac{[a^2+b^2]}{a^2} \\ =\left[\frac{1}{{\cos }^2\alpha }\right] \\ =\frac{1}{\cos \alpha } \end{gathered}$$

Foci are $ae=\cos \alpha \times \left(\frac{1}{\cos \alpha }\right)=1$

So foci does not changes

Hence, the abscissae foci remains constant with the change in $‘\alpha ’$

Explanation for the in-correct option:

Option A :

Vertices are $(\pm a,0)=(\pm {\cos }^2\alpha ,0)$

Option C:

Eccentricity $e=\frac{1}{\cos \alpha }$

Option D:

Directrix

$x=\pm \frac{a}{e}=\frac{{\cos }^2\alpha }{{\displaystyle \frac{1}{\cos \alpha }}}={\cos }^3\alpha$

Hence, option (B) is correct option