Question

For the hyperbola $\frac{x^2}{co{s}^2\alpha }-\frac{y^2}{si{n}^2\alpha }=1,$ when $\alpha$ changes, remains constant.

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

Answer 1

The correct option is B abscissa of foci

Given hyperbola is,

$\frac{x^2}{{\cos }^2\alpha }-\frac{y^2}{{\sin }^2\alpha }=1$

Comparing with a standard hyperbola we get,

$a^2={\cos }^2\alpha$

$b^2={\sin }^2\alpha$

We are asked to find which of the given options remain unchanged with change in $\alpha$.

For finding that, we need to see which of the quantities remain independent of $\alpha$.

$b^2=a^2(e^2-1)$

$e^2-1=\frac{b^2}{a^2}$

$=\frac{si{n}^2\propto }{co{s}^2\propto }=ta{n}^2\propto$

$e=|\sec \alpha |$

directrix,$x=\pm \frac{a}{e}=\pm \frac{\cos \alpha }{\sec \alpha }$

Abscissa of focus $=\pm ae$

$=\pm \cos \alpha |\sec \alpha |$

$=\pm 1$ which is independent of $\alpha$