If the angle between the asymptotes of the hy | vedclass

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

Question

(C) The given hyperbola is $\frac{x^2}{4a^2} - \frac{y^2}{12} = 1$.Here,$A^2 = 4a^2$ and $B^2 = 12$,so $A = 2a$ and $B = 2\sqrt{3}$.The angle $	heta$

Vedclass · Accepted Answer

$\frac{y^2}{12} - \frac{x^2}{36} = 1$

Vedclass · Answer

(C) The given hyperbola is $\frac{x^2}{4a^2} - \frac{y^2}{12} = 1$.Here,$A^2 = 4a^2$ and $B^2 = 12$,so $A = 2a$ and $B = 2\sqrt{3}$.The angle $	heta$ between the asymptotes is given by $2 	an^{-1}(\frac{B}{A}) = \frac{\pi}{3}$.Thus,$	an^{-1}(\frac{B}{A}) = \frac{\pi}{6}$,which implies $\frac{B}{A} = 	an(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$.Substituting the values,$\frac{2\sqrt{3}}{2a} = \frac{1}{\sqrt{3}}$ $\Rightarrow \frac{\sqrt{3}}{a} = \frac{1}{\sqrt{3}}$ $\Rightarrow a = 3$.Then $A^2 = 4(3)^2 = 36$.The equation of the hyperbola is $\frac{x^2}{36} - \frac{y^2}{12} = 1$.The conjugate hyperbola is $\frac{x^2}{36} - \frac{y^2}{12} = -1$,which is $\frac{y^2}{12} - \frac{x^2}{36} = 1$.

If the angle between the asymptotes of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{3} = 4$ is $\frac{\pi}{3}$,then its conjugate hyperbola is:

Similar Questions

The length of the latus rectum of the curve $xy = 7x + 5y$ is

The length of the conjugate axis of a hyperbola is greater than the length of the transverse axis. Then, the eccentricity $e$ is

The equation of the pair of asymptotes of the hyperbola $x^2-2 y^2-8 x+8 y+4=0$ is

The equation of the chord of the hyperbola $25x^{2} - 16y^{2} = 400$ whose midpoint is $(5, 3)$ is:

The point $P(-2 \sqrt{6}, \sqrt{3})$ lies on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ having eccentricity $e = \frac{\sqrt{5}}{2}$. If the tangent and normal at $P$ to the hyperbola intersect its conjugate axis at the points $Q$ and $R$ respectively,then $QR$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module