For the hyperbola frac{x^{2}}{cos^{2} alpha} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The given equation of the hyperbola is $\frac{x^{2}}{\cos^{2} \alpha} - \frac{y^{2}}{\sin^{2} \alpha} = 1$. Comparing this with the standard equat

Vedclass · Accepted Answer

Foci

Vedclass · Answer

(C) The given equation of the hyperbola is $\frac{x^{2}}{\cos^{2} \alpha} - \frac{y^{2}}{\sin^{2} \alpha} = 1$. Comparing this with the standard equation $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$, we get $a^{2} = \cos^{2} \alpha$ and $b^{2} = \sin^{2} \alpha$. The coordinates of the foci are given by $(\pm ae, 0)$, where $ae = \sqrt{a^{2} + b^{2}}$. Substituting the values, $ae = \sqrt{\cos^{2} \alpha + \sin^{2} \alpha} = \sqrt{1} = 1$. Thus, the foci are $(\pm 1, 0)$, which are independent of $\alpha$. Therefore, the foci remain fixed as $\alpha$ varies.

For the hyperbola $\frac{x^{2}}{\cos^{2} \alpha} - \frac{y^{2}}{\sin^{2} \alpha} = 1$, which of the following remains fixed when $\alpha$ varies?

Similar Questions

The product of the perpendicular distances from any point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ to its asymptotes is

If the length of the latus rectum of a hyperbola is $8$ and its eccentricity is $\frac{3}{\sqrt{5}}$,then the equation of the hyperbola is:

The equation of the pair of asymptotes of the hyperbola $4x^2 - 9y^2 - 24x - 36y - 36 = 0$ is

Find the equation of the hyperbola satisfying the given conditions: Foci $(\pm 4, 0)$,the latus rectum is of length $12$.

Which of the following equations in parametric form can represent a hyperbola,where $t$ is a parameter?

Vedclass Test Series

Exam Paper Generator

Online Exam Module