Chapter EP, Problem 10.5.9EP

To determine

To find: The coordinates of the vertices, foci, and asymptotes. And hence graph the hyperbola.

Answer to Problem 10.5.9EP

The coordinates of the vertices, foci, and asymptotes of the hyperbola are given below.

Explanation of Solution

Given:

The given equation of the hyperbola is as:

  $x^2+10x+5=4y^2+16$

Rewrite the given equation as:

  $\begin{array}{c}{x}^2+10x+25-25-4y^2=11\\ {\left(x+5\right)}^2-4y^2=36\\ \frac{{\left(x+5\right)}^2}{36}-\frac{y^2}{9}=1\end{array}$

Consider the standard equation of the hyperbola with centre at $\left(h,k\right)$ , as:

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

By comparing the given equation of hyperbola with the standard equation of the hyperbola, the centre is given as:

  $\left(h,k\right)=\left(-5,0\right)$

The value of $a^2$ and $b^2$ is as:

  $\begin{array}{c}{a}^2=36\\ a=6\\ b^2=9\\ b=3\end{array}$

The focal length $c$ is calculated as:

  $\begin{array}{c}c=\sqrt{a^2+b^2}\\ =\sqrt{36+9}\\ =\sqrt{45}\\ \approx 6.708\end{array}$

The coordinate of foci is given as:

  $\left(h\pm c,k\right)=\left(-5\pm 6.708,0\right)$ .

So the coordinate of the foci are as:

  $\left(1.708,0\right),\left(-11.708,0\right)$ .

The vertices of the hyperbola is given as:

  $\begin{array}{c}\left(h\pm a,k\right)=\left(-5\pm 6,0\right)\\ =\left(-11,0\right)\&\left(1,0\right)\end{array}$ .

The equation of asymptotes is given as:

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

So the equation of asymptotes of given equation is as:

  $\begin{array}{c}y-\left(0\right)=\pm \frac{3}{6}\left(x+5\right)\\ y=\pm \frac{1}{2}\left(x+5\right)\end{array}$

The graph of the given hyperbola is given below: