Chapter 8.6, Problem 6GP

Solution

To write: the equation of the hyperbola.

  $\frac{{\left(x+4\right)}^2}{9}-\frac{{\left(y-3\right)}^2}{16}=1$

Given information:

The vertices of the hyperbola are at $\left(-7,3\right)$ and $\left(-1,3\right)$ and foci at $\left(-9,3\right)$ and $\left(1,3\right)$ .

Formula Used:

Standard equation of an Hyperbola

Equation	Transverse Axis	Vertices	Asymptotes
$\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$	Horizontal	$\left(h\pm a,k\right)$	$y=\pm \frac{b}{a}\left(x-h\right)+k$
$\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1$	Vertical	$\left(h,k\pm a\right)$	$y=\pm \frac{a}{b}\left(x-h\right)+k$

The foci of the hyperbola lie on the transverse axis at a distance of c units from the center, where $c^2=a^2+b^2$

Explanation:

Center of the hyperbola is at the middle of the two vertices.

Since, the y co-ordinates are same, so it has horizontal transverse axis and has the form:

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

The center is between -7 and -1.

  $\begin{array}{c}h=\frac{-1+\left(-7\right)}{2}\\ =\frac{-1-7}{2}\\ =-\frac{8}{2}\\ =-4\end{array}$

So, the center is at $\left(h,k\right)=\left(-4,3\right)$ .

Distance from center to vertex a :

  $\begin{array}{c}a=\left|-7-h\right|\\ =\left|-7-\left(-4\right)\right|\\ =\left|-7+4\right|\\ =\left|-3\right|\\ =3\end{array}$

The foci of hyperbola is at $\left(-9,3\right)$ and $\left(1,3\right)$ . Distance from center to focus is:

  $\begin{array}{c}c=\left|-9-h\right|\\ =\left|-9-\left(-4\right)\right|\\ =\left|-9+4\right|\\ =\left|-5\right|\\ =5\end{array}$

Now, find the distance from the center to co-vertex b .

  $\begin{array}{c}{c}^2=a^2+b^2\\ 5^2=3^2+b^2\\ 25=9+b^2\\ b^2=25-9\\ b^2=16\\ b=4\end{array}$

Now susbtitute these values in the standard equation of hyperbola.

  $\begin{array}{c}\frac{{\left(x-\left(-4\right)\right)}^2}{3^2}-\frac{{\left(y-3\right)}^2}{4^2}=1\\ \frac{{\left(x+4\right)}^2}{9}-\frac{{\left(y-3\right)}^2}{16}=1\end{array}$