Chapter 8.3, Problem 35E

Solution

To calculate: To determine an equation in standard form of the hyperbola.

The standard form of an equation is:

  $\frac{{\left(x+1\right)}^2}{4}-\frac{{\left(y-2\right)}^2}{5}=1$

Given information: The given condition of a hyperbola is:

Foci $\left(-4,2\right),\left(2,2\right)$ , transverse axis endpoints are $\left(-3,2\right),\left(1,2\right)$ .

Calculation:

In a hyperbola, there are two vertices, two foci, transverse axis and conjugate axis.

The midpoint of vertices is at the center of the hyperbola. As there is common $y$ coordinate in vertices, hyperbola is horizontal.

The general form of hyperbola is:

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

The distance between two vertices is $2a$ , distance between two foci is $2c$ , distance between two co-vertices is $2b$ .

Now, the values can be calculated as:

  $\begin{array}{c}\Rightarrow 2a=\sqrt{{\left(1-(-3\right)}^2+{\left(2-2\right)}^2}\\ \Rightarrow 2a=\sqrt{{\left(1+3\right)}^2+0}\\ \Rightarrow 2a=4\\ \Rightarrow a=2\end{array}$

  $\begin{array}{c}\Rightarrow 2c=\sqrt{{\left(2-(-4\right)}^2+{\left(2-2\right)}^2}\\ \Rightarrow 2c=\sqrt{{\left(2+4\right)}^2+0}\\ \Rightarrow 2c=6\\ \Rightarrow c=3\end{array}$

The value of

  $c=\sqrt{a^2+b^2}$ .

  $\begin{array}{c}{3}^2=2^2+b^2\\ 9=4+b^2\\ 9-4=b^2\\ 5=b^2\\ b=\pm \sqrt{5}\end{array}$

Thus, the equation of a hyperbola is,

  $\begin{array}{l}\frac{{\left(x-\left(-1\right)\right)}^2}{2^2}-\frac{{\left(y-2\right)}^2}{{\left(\sqrt{5}\right)}^2}=1\\ \frac{{\left(x+1\right)}^2}{4}-\frac{{\left(y-2\right)}^2}{5}=1\end{array}$