Chapter 10.4, Problem 33E

To determine

To find:The centre, vertices, foci, and equations of the asymptotes of the hyperbola and the sketch of the hyperbola.

Answer to Problem 33E

The centre, vertices, foci, and equations of the asymptotes of the hyperbola are $\left(0,0\right)$ , $\left(\sqrt{3},0\right)$ and $\left(-\sqrt{3},0\right)$ , $\left(\pm \sqrt{5},0\right)$ , and $y=\pm \sqrt{\frac{2}{3}}x$ and the sketch of the hyperbola is shown in figure (1).

Explanation of Solution

Given information:

The equation is $2x^2-3y^2=6$ .

Calculation:

The standard form of the equation is,

  $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$   ...... (1)

Rearrange the above equation.

  $\frac{{\left(x-0\right)}^2}{{\left(\sqrt{3}\right)}^2}-\frac{{\left(y-0\right)}^2}{{\left(\sqrt{2}\right)}^2}=1$   ...... (2)

Compare equation (1) and (2).

  $\begin{array}{l}\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=\frac{{\left(x-0\right)}^2}{{\left(\sqrt{3}\right)}^2}-\frac{{\left(y-0\right)}^2}{{\left(\sqrt{2}\right)}^2}\\ a=\sqrt{3}\\ b=\sqrt{2}\end{array}$

Therefore, the center is at $\left(0,0\right)$ .

The transverse axis of the hyperbola is horizontal.

Calculate the value of $c$ .

  $\begin{array}{c}c=\sqrt{a^2+b^2}\\ =\sqrt{{\left(\sqrt{3}\right)}^2+{\left(\sqrt{2}\right)}^2}\\ =\sqrt{5}\end{array}$

The vertices and foci are given as $\left(\pm a+h,k\right)$ and $\left(\pm c+h,k\right)$ respectively. Hence the vertices as $\left(\sqrt{3},0\right)$ and $\left(-\sqrt{3},0\right)$ , and the foci is $\left(\pm \sqrt{5},0\right)$ .

The equation of the asymptotes of the hyperbola is,

  $y=k\pm \frac{b}{a}\left(x-h\right)$

Therefore, the equations of the asymptotes are $y=\pm \sqrt{\frac{2}{3}}x$

Draw the sketch for the hyperbola by using the equations of asymptotes.

  

Figure-(1)

Therefore, the centre, vertices, foci, and equations of the asymptotes of the hyperbola are $\left(0,0\right)$ , $\left(\sqrt{3},0\right)$ and $\left(-\sqrt{3},0\right)$ , $\left(\pm \sqrt{5},0\right)$ , and $y=\pm \sqrt{\frac{2}{3}}x$ and the sketch of the hyperbola is shown in figure (1).