# How do you find the center, vertices, foci and asymptote of 2x^2-3y^2=6?

Feb 28, 2018

Centre $\left(0 , 0\right)$, Vertices are $\left(- \sqrt{3} , 0\right) \mathmr{and} \left(\sqrt{3} , 0\right)$,
Focii are $\left(- \sqrt{5} , 0\right) \mathmr{and} \left(\sqrt{5} , 0\right)$ Asymptote is
$y = \pm \sqrt{\frac{2}{3}} x$

#### Explanation:

$2 {x}^{2} - 3 {y}^{2} = 6 \mathmr{and} \frac{2 {x}^{2}}{6} - \frac{3 {y}^{2}}{6} = 1$ or

${x}^{2} / 3 - {y}^{2} / 2 = 1 \mathmr{and} {\left(x - 0\right)}^{2} / {\left(\sqrt{3}\right)}^{2} - {\left(y - 0\right)}^{2} / {\left(\sqrt{2}\right)}^{2} = 1$

This is standard form of the equation of a hyperbola with center

(h,k) ; (x-h)^2/a^2-(y-k)^2/b^2=1 :  Centre $\left(0 , 0\right)$

The vertices are $a$ units from the center, and the foci are

$c$ units from the center. Moreover ${c}^{2} = {a}^{2} + {b}^{2}$

Vertices are (-a,0) and (a,0) ; a = sqrt3 and b=sqrt2 or

(-sqrt3,0) and (sqrt3,0) ; c^2= (sqrt3)^2+sqrt(2)^2=5

$\therefore c = \pm \sqrt{5} \therefore$ Focii are at $\left(- \sqrt{5} , 0\right) \mathmr{and} \left(\sqrt{5} , 0\right)$

Each hyperbola has two asymptotes that intersect at the center of

the hyperbola. The asymptotes pass through the vertices of a

rectangle of dimensions $2 a = 2 \sqrt{3} \mathmr{and} 2 b = 2 \sqrt{2}$ with its

centre at $\left(0 , 0\right) \therefore$ slope $= \pm \left(\frac{\sqrt{2}}{\sqrt{3}}\right)$ Equation of

asymptote is $y - 0 = \pm \sqrt{\frac{2}{3}} \left(x - 0\right) \mathmr{and} y = \pm \sqrt{\frac{2}{3}} x$

graph{2x^2-3y^2=6 [-10, 10, -5, 5]} [Ans]