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

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

Question

(B) The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,where $a^2 = \cos^2 \alpha$ and $b^2 = \sin^2 \alpha$.For a hyperbola,the

Vedclass · Accepted Answer

Abscissae of foci

Vedclass · Answer

(B) The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,where $a^2 = \cos^2 \alpha$ and $b^2 = \sin^2 \alpha$.For a hyperbola,the foci are located at $(\pm ae, 0)$,where $e$ is the eccentricity.The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{\sin^2 \alpha}{\cos^2 \alpha}} = \sqrt{1 + 	an^2 \alpha} = \sec \alpha$.The abscissa of the foci is $\pm ae = \pm \sqrt{\cos^2 \alpha} \cdot \sec \alpha = \pm \cos \alpha \cdot \frac{1}{\cos \alpha} = \pm 1$.Since the value $\pm 1$ is independent of $\alpha$,the abscissae of the foci remain constant.

For the hyperbola $\frac{x^2}{\cos^2 \alpha} - \frac{y^2}{\sin^2 \alpha} = 1$,which of the following remains constant with a change in $\alpha$?

Similar Questions

If $L_1=0$ and $L_2=0$ are the asymptotes of the hyperbola $9x^2-4y^2+36x+8y-4=0$,then the product of the perpendicular distances from the point $(1,1)$ to the lines $L_1=0$ and $L_2=0$ is

The equation of the normal to the hyperbola $\frac{x^2}{25} - \frac{y^2}{16} = 1$ that is perpendicular to the line $2x + y = 1$ is:

$A$ hyperbola having its centre at the origin passes through the point $(5, 2)$ and has a transverse axis of length $8$ along the $X$-axis. What is the eccentricity of its conjugate hyperbola?

The line $3x - 4y = 5$ is a tangent to the hyperbola $x^2 - 4y^2 = 5$. The point of contact is

What kind of hyperbola does the equation $9x^2 - 16y^2 - 18x + 32y - 151 = 0$ represent?

Vedclass Test Series

Exam Paper Generator

Online Exam Module