Chapter 10.4, Problem 34E

To determine

To sketch the graph of the hyperbolic equation $6y^2-3x^2=18$ and its asymptotes and hence find the centre, vertices, foci.

Answer to Problem 34E

  $centre=\left(h,k\right)=\left(0,0\right),vertices=\left(0,\sqrt{3}\right)and\left(0,-\sqrt{3}\right),foci=\left(0,\pm 3\right)$ and the asymptotes are $y=\pm \sqrt{\frac{1}{2}}x$ .

Explanation of Solution

Given information:

Given equation of hyperbola $6y^2-3x^2=18$

Since the coefficient of $y^2$ is positive, so the hyperbola has a vertical transversal axis.

And also given $6y^2-3x^2=18$

  $\Rightarrow \frac{y^2}{3}-\frac{x^2}{6}=1$

Compare given hyperbola equation $\frac{y^2}{3}-\frac{x^2}{6}=1$ with standard form of hyperbola equation $\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b}=1$ , we get

  $\begin{array}{l}centre=\left(h,k\right)=\left(0,0\right),a^2=3,b^2=6.\\ \Rightarrow a=\sqrt{3},b=\sqrt{6}\end{array}$

As we know that $c^2=a^2+b^2$

  $\begin{array}{l}\Rightarrow c^2=3+6\Rightarrow c^2=9\\ \therefore c=\pm 3\end{array}$

  $\begin{array}{l}\text{Thus,}\text{vertices}=\left(h,k\pm a\right)=\left(0,0\pm \sqrt{3}\right)=\left(0,\pm \sqrt{3}\right)\\ Foci=\left(h,k\pm c\right)=\left(0,0\pm 3\right)=\left(0,\pm 3\right)\end{array}$

Since hyperbola is vertical, the asymptotes equations are $y-k=\pm \frac{a}{b}\left(x-h\right)$

  $\begin{array}{l}\Rightarrow y=\pm \frac{\sqrt{3}}{\sqrt{6}}\left(x\right)\\ \Rightarrow y=\pm \sqrt{\frac{1}{2}}x\end{array}$

Graph:

The graph of hyperbola equation $\frac{y^2}{3}-\frac{x^2}{6}=1$ with its asymptotes

  

Interpretation: From above graph, we observe that graph satisfies has a vertical transverse axis with centre $\left(h,k\right)=\left(0,0\right)$ .

Hence, $centre=\left(h,k\right)=\left(0,0\right),vertices=\left(0,\sqrt{3}\right)and\left(0,-\sqrt{3}\right),foci=\left(0,\pm 3\right)$ and the asymptotes are $y=\pm \sqrt{\frac{1}{2}}x$ .