Chapter 12.3, Problem 49E

To determine

To calculate:

The coordinates of the center, foci, vertices and also the equation of the asymptotes.

Answer to Problem 49E

The center is $\left(6,-2\right)$ , foci are $\left(6+\sqrt{5},-2\right)$ and $\left(6-\sqrt{5},-2\right)$ ,the vertices are $\left(8,-2\right)$ and $\left(4,-2\right)$ , asymptotes are $y=-\frac{1}{2}x-1$ and $y=\frac{1}{2}x+5$ .

Explanation of Solution

Formula used:

  $a_n=a_1+\left(n-1\right)d$

Calculation:

Consider the hyperbola equation $x^2-4y^2-12x-16y=-16$ .

Thus, to name the coordinates of the center, foci, and vertices, and the equation of the asymptotes, let’s write the equation in standard form as:

  $\begin{array}{l}{x}^2-12x-4\left(y^2+4y\right)=-16\\ x^2-12x+36-4\left(y^2+4y+4\right)=-16+36-16\\ {\left(x-6\right)}^2-4{\left(y+2\right)}^2=36-32\\ {\left(x-6\right)}^2-4{\left(y+2\right)}^2=4\\ \frac{{\left(x-6\right)}^2}{4}-\frac{{\left(y+2\right)}^2}{1}=1\end{array}$

The equation is the form of $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$ where center is $\left(h,k\right)$ , foci is $\left(h\pm c,k\right)$ , vertices is $\left(h\pm a,k\right)$ and equation of transverse axis is $y=k$ .

The center $\left(h,k\right)$ is $\left(6,-2\right)$ .

The $x$ terms are in the first expression; the transverse axis is horizontal.

  $a=2,b=1$

  $\begin{array}{l}{c}^2=a^2+b^2\\ c^2=4+1\\ c^2=5\\ c=\sqrt{5}\end{array}$

The foci $\left(h\pm c,k\right)$ are $\left(6+\sqrt{5},-2\right)$ and $\left(6-\sqrt{5},-2\right)$

The vertices $\left(h\pm a,k\right)$ are $\left(8,-2\right)$ and $\left(4,-2\right)$ .

Equations of asymptotes of hyperbola is:

  $\begin{array}{l}y-k=\pm \frac{b}{a}\left(x-h\right)\\ y-\left(-2\right)=\pm \frac{1}{2}\left(x-6\right)\\ y+2=\pm \frac{1}{2}x\pm 3\\ y+2=-\frac{1}{2}x-3andy+2=\frac{1}{2}x+3\\ y=-\frac{1}{2}x-1andy=\frac{1}{2}x+5\end{array}$

Therefore, the center is $\left(6,-2\right)$ , foci are $\left(6+\sqrt{5},-2\right)$ and $\left(6-\sqrt{5},-2\right)$ ,the vertices are $\left(8,-2\right)$ and $\left(4,-2\right)$ , asymptotes are $y=-\frac{1}{2}x-1$ and $y=\frac{1}{2}x+5$ .