Question

For the hyperbola $\frac{x^2}{co{s}^2\alpha }-\frac{y^2}{si{n}^2\alpha }=1$

Which one of the following remain constant with change of $\alpha$ ?

• 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,

$\frac{x^2}{co{s}^2\propto }-\frac{y^2}{si{n}^2\propto }=1$

Comparing with a standard hyperbola we get,

$a^2=co{s}^2\alpha$

$b^2=si{n}^2\alpha$

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 $\alpha$.

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

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

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

$e=|sec\propto |$

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

Abscissa of focus=$\pm ae$

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

$=\pm 1$

Which is independent of $\alpha$.

Hence answer (b).