Chapter 9.3, Problem 30E

To determine

To find: The centre, vertices, foci, and asymptotes of the hyperbola, and sketch its graph using the asymptotes as an aid.

Answer to Problem 30E

The centre is $\left(-3,1\right)$ , vertices are $\left(-3,1\pm \frac{1}{2}\right)$ , foci are $\left(-3,1\pm \frac{2\sqrt{5}}{8}\right)$ and the asymptotes are $y=-1\pm 2\left(x+3\right)$ .

Explanation of Solution

Given information:

The given hyperbola is $\frac{{\left(y-1\right)}^2}{1/4}-\frac{{\left(x+3\right)}^2}{1/16}=1$

Calculation:

The given hyperbola is

  $\frac{{\left(y-1\right)}^2}{1/4}-\frac{{\left(x+3\right)}^2}{1/16}=1$

From the above equation it can be concluded that the hyperbola has a vertical transverse axis.

The centre of the hyperbola is $\left(h,k\right)=\left(-3,1\right)$ .

The vertices of the hyperbola is $(h,k\pm a)=\left(-3,1\pm \frac{1}{2}\right)$ , that is $\left(-3,\frac{3}{2}\right)$ and $\left(-3,\frac{1}{2}\right)$ and the conjugate axis with end point $(h\pm b,k)=\left(-3\pm \frac{1}{4},1\right)$ , that is $\left(-\frac{11}{2},1\right)$ and $\left(-\frac{13}{2},1\right)$ . To sketch the graph draw a rectangle through these four point. The asymptotes are the lines that passes through the corner of the rectangle.

The equation of the asymptotes are,

  $\begin{array}{l}y=k\pm \frac{a}{b}\left(x-h\right)\\ y=1\pm \frac{\frac{1}{2}}{\frac{1}{4}}\left(x+3\right)\\ y=-1\pm 2\left(x+3\right)\end{array}$

Now, find the foci by using the equation $c^2=a^2+b^2$ , therefore,

  $\begin{array}{l}{c}^2=\frac{1}{4}+\frac{1}{16}\\ c^2=\frac{4+16}{64}\\ c=\sqrt{\frac{20}{64}}\\ c=\frac{2\sqrt{5}}{8}\end{array}$

Thus, the foci are $\left(h,k\pm c\right)=\left(-3,1\pm \frac{2\sqrt{5}}{8}\right)$ .

The sketch of the hyperbola is drawn below:

  

Now, verify the graph by graphing utility:

  

Hence,

The centre is $\left(-3,1\right)$ , vertices are $\left(-3,1\pm \frac{1}{2}\right)$ , foci are $\left(-3,1\pm \frac{2\sqrt{5}}{8}\right)$ and the asymptotes are $y=-1\pm 2\left(x+3\right)$ .