The angle between the asymptotes of the hyper | 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 $x^2-3y^2=3$. Dividing by $3$,we get $\frac{x^2}{3}-\frac{y^2}{1}=1$.Here,$a^2=3$ and $b^2=1$,so $a=\sqrt{3

Vedclass · Accepted Answer

$\frac{\pi}{3}$

Vedclass · Answer

(C) The given equation of the hyperbola is $x^2-3y^2=3$. Dividing by $3$,we get $\frac{x^2}{3}-\frac{y^2}{1}=1$.Here,$a^2=3$ and $b^2=1$,so $a=\sqrt{3}$ and $b=1$.The equations of the asymptotes are $y = \pm \frac{b}{a}x$,which gives $y = \frac{1}{\sqrt{3}}x$ and $y = -\frac{1}{\sqrt{3}}x$.Let the slopes be $m_1 = \frac{1}{\sqrt{3}}$ and $m_2 = -\frac{1}{\sqrt{3}}$.The angle $	heta$ between the asymptotes is given by $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$.Substituting the values,$	an 	heta = \left| \frac{\frac{1}{\sqrt{3}} - (-\frac{1}{\sqrt{3}})}{1 + (\frac{1}{\sqrt{3}})(-\frac{1}{\sqrt{3}})} ight| = \left| \frac{\frac{2}{\sqrt{3}}}{1 - \frac{1}{3}} ight| = \left| \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}} ight| = \sqrt{3}$.Since $	an 	heta = \sqrt{3}$,we have $	heta = \frac{\pi}{3}$.

The angle between the asymptotes of the hyperbola $x^2-3y^2=3$ is

Similar Questions

Let $L(ae, b^2/a)$ be the end of the latus rectum of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ lying in the first quadrant,and let $S(ae, 0)$ be the focus of the given hyperbola. Given $L$ is $(x_1, 4)$ and $S$ is $(8, y_1)$,find the length of its transverse axis.

The locus of the midpoint of the system of parallel chords parallel to the line $y = 2x$ for the hyperbola $9x^2 - 4y^2 = 36$ is

The equation of the tangents to the conic $3x^2 - y^2 = 3$ perpendicular to the line $x + 3y = 2$ is

The equations of the tangents to the hyperbola $4x^2 - y^2 = 12$ are $y = 4x + c_1$ and $y = 4x + c_2$. Then $|c_1 - c_2|$ is equal to -

The equation $\frac{1}{r} = \frac{1}{8} + \frac{3}{8} \cos \theta$ represents:

Vedclass Test Series

Exam Paper Generator

Online Exam Module