Chapter 8.7, Problem 10P

Solution

To write: the equation of the hyperbola.

  $\frac{25{\left(y+1\right)}^2}{64}-\frac{25{\left(x-4\right)}^2}{336}=1$

Given information:

The foci of the hyperbola is at $\left(4,-5\right)$ and $\left(4,3\right)$ and $e=2.5$

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 $c^2=a^2+b^2$

Eccentricity of conic sections

The eccentricity of each conic section is defined below. For an ellipse or hyperbola, c is the distance from each focus to the center, and a is the distance from each vertex to center.

Circle: $e=0$

Parabola: $e=1$

Ellipse: $e=\frac{c}{a},\text{ and }0<e<1$

Hyperbola: $e=\frac{c}{a},\text{ and }e>1$

Explanation:

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

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

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

The center is between -5 and 3.

  $\begin{array}{c}k=\frac{-5+3}{2}\\ =-\frac{2}{2}\\ =-1\end{array}$

So, the center is at $\left(h,k\right)=\left(4,-1\right)$ .

The foci of hyperbola is at $\left(4,-5\right)$ and $\left(4,3\right)$ . Distance from center to focus is:

  $\begin{array}{c}c=3-k\\ =3-\left(-1\right)\\ =3+1\\ =4\end{array}$

It is given that the eccentricity of the given hyperbola is:

  $\begin{array}{c}e=2.5\\ \frac{c}{a}=\frac{25}{10}\\ \frac{4}{a}=\frac{5}{2}\\ a=\frac{8}{5}\end{array}$

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

  $\begin{array}{c}{c}^2=a^2+b^2\\ 4^2={\left(\frac{8}{5}\right)}^2+b^2\\ 16=\frac{64}{25}+b^2\\ b^2=16-\frac{64}{25}\\ b^2=\frac{400-64}{25}\\ b^2=\frac{336}{25}\end{array}$

Now susbtitute these values in the standard equation of hyperbola.

  $\begin{array}{c}\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1\\ \frac{{\left(y-\left(-1\right)\right)}^2}{\frac{64}{25}}-\frac{{\left(x-4\right)}^2}{\frac{336}{25}}=1\\ \frac{25{\left(y+1\right)}^2}{64}-\frac{25{\left(x-4\right)}^2}{336}=1\end{array}$