Chapter 10.4, Problem 20E

To determine

To find the centre, vertices, foci and the equation of the asymptotes of the hyperbola $\frac{x^2}{9}-\frac{y^2}{16}=1$ and then sketch the hyperbola using its asymptotes.

Answer to Problem 20E

  $centre=\left(h,k\right)=\left(0,0\right),\text{vertices}=\left(\pm 3,0\right),foci=\left(\pm 5,0\right)$ and the asymptotes are $y=\pm \frac{4}{3}x$

Explanation of Solution

Given information :

Given equation of hyperbola $\frac{x^2}{9}-\frac{y^2}{16}=1$

Since the coefficient of $x^2$ is positive, so the hyperbola has a horizontal transversal axis. Compare given hyperbola equation $\frac{x^2}{9}-\frac{y^2}{16}=1$ with standard form of hyperbola equation $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b}=1$ , we get

  $\begin{array}{l}centre=\left(h,k\right)=\left(0,0\right),a^2=9,b^2=16\\ \Rightarrow a=3,b=4\end{array}$ .

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

  $\begin{array}{l}\Rightarrow c^2=9+16\Rightarrow c^2=25\\ \therefore c=\pm 5\end{array}$

  $\begin{array}{l}\text{Thus,}\text{vertices}=\left(h\pm a,k\right)=\left(0\pm 3,0\right)=\left(\pm 3,0\right)\\ Foci=\left(h\pm c,k\right)=\left(0\pm 5,0\right)=\left(\pm 5,0\right)\end{array}$

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

  $\begin{array}{l}\Rightarrow y=\pm \frac{4}{3}\left(x-0\right)\\ \therefore y=\pm \frac{4}{3}x\end{array}$

Graph:

The graph of hyperbola equation $\frac{x^2}{9}-\frac{y^2}{16}=1$ with its asymptotes

  

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

Hence, $centre=\left(h,k\right)=\left(0,0\right),\text{vertices}=\left(\pm 3,0\right),foci=\left(\pm 5,0\right)$ and the asymptotes are $y=\pm \frac{4}{3}x$