Chapter 9.3, Problem 19E

To determine

To find: The standard form of the equation of the hyperbola with the given characteristics.

Answer to Problem 19E

The standard form of the equation of the hyperbola is $\frac{{\left(x-2\right)}^2}{4}-\frac{{\left(y+3\right)}^2}{8}=1$ .

Explanation of Solution

Given information:

Vertices: $\left(0,-3\right)$ , $\left(4,-3\right)$ and passes through the point $\left(-4,5\right)$ .

Calculation:

The standard form of the equation of the hyperbola is with horizontal transverse axis is

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

where $\left(h,k\right)$ is the centre of the hyperbola and a is the distance the vertices from the centre of hyperbola.

Since, the vertices are $\left(0,-3\right)$ and $\left(4,-3\right)$ , therefore by mid-point formula the centre of the hyperbola occur at the point

  $\left(\frac{0+4}{2},\frac{-3+\left(-3\right)}{2}\right)=\left(2,-3\right)$

So, $\left(h,k\right)=\left(2,-3\right)$

Also,

  $\begin{array}{l}2a=0+4\\ 2a=4\\ a=2\end{array}$

Substitute $\left(h,k\right)=\left(2,-3\right)$ and $a=2$ in equation (1):

  $\begin{array}{l}\frac{{\left(x-2\right)}^2}{2^2}-\frac{{\left(y-\left(-3\right)\right)}^2}{b^2}=1\\ \frac{{\left(x-2\right)}^2}{4}-\frac{{\left(y+3\right)}^2}{b^2}=1\end{array}$......(2)

Since, the hyperbola passes through the point $\left(-4,5\right)$ , therefore substitute this point in equation (ii) and solve for b:

  $\begin{array}{l}\frac{{\left(x-2\right)}^2}{4}-\frac{{\left(y+3\right)}^2}{b^2}=1\\ \frac{{\left(-4-2\right)}^2}{4}-\frac{{\left(5+3\right)}^2}{b^2}=1\\ \frac{36}{4}-\frac{64}{b^2}=1\\ 9-1=\frac{64}{b^2}\\ 8=\frac{64}{b^2}\\ b^2=\frac{64}{8}\\ b^2=8\end{array}$

Substitute $b^2=8$ in equation (ii) and solve:

  $\frac{{\left(x-2\right)}^2}{4}-\frac{{\left(y+3\right)}^2}{8}=1$

Hence,

The standard form of the equation of the hyperbola is $\frac{{\left(x-2\right)}^2}{4}-\frac{{\left(y+3\right)}^2}{8}=1$ .