Chapter 9.3, Problem 28E

To determine

To find: The center, vertices, foci, asymptotes and graph of hyperbola $\frac{{\left(x-2\right)}^2}{4}-\frac{{\left(y+5\right)}^2}{25}=1$ . Verify the result using graphical utility.

Answer to Problem 28E

The center of the hyperbola is $\left(2,-5\right)$ , the vertices of the hyperbola is $\left(4,-5\right)$ , $\left(0,-5\right)$ , the foci of the hyperbola is $\left(2+\sqrt{29},-5\right)$ and $\left(2-\sqrt{29},-5\right)$ , the asymptotes of the hyperbola is $y=\frac{5x}{2}-10$ , $y=-\frac{5x}{2}$ and the graph of the hyperbola is shown in figure (1).

Explanation of Solution

Given information:

Given equation of hyperbola is $\frac{{\left(x-2\right)}^2}{4}-\frac{{\left(y+5\right)}^2}{25}=1$ .

Calculation:

Compare the given equation with the standard equation of hyperbola $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$ .

  $\begin{array}{l}a=2\\ b=5\\ h=2\\ k=-5\end{array}$

The center of the hyperbola is given as $\left(2,-5\right)$ .

The vertices of the hyperbola is given by $\left(h+a,k\right)$ and $\left(h-a,k\right)$ .

So the vertices are $\left(4,-5\right)$ and $\left(0,-5\right)$ .

The foci of the hyperbola is given by $\left(h+c,k\right)$ and $\left(h-c,k\right)$ , where $c^2=a^2+b^2$ . Calculate the value of $c$ .

  $\begin{array}{c}{c}^2=2^2+5^2\\ c^2=4+25\\ c^2=29\\ c=\sqrt{29}\end{array}$

Hence the foci are $\left(2+\sqrt{29},-5\right)$ and $\left(2-\sqrt{29},-5\right)$ .

The asymptotes of the hyperbola is given by $y=\pm \frac{b}{a}\left(x-h\right)+k$ .

Substitute $2$ for $a$ and $-5$ for $b$ , $2$ for h and $-5$ for k to find the asymptotes.

  $y=\frac{5x}{2}-10$ , $y=-\frac{5x}{2}$

To draw the graph of hyperbola, solve the equations of hyperbola in terms of $y$

  $\begin{array}{c}\frac{{\left(x-2\right)}^2}{4}-\frac{{\left(y+5\right)}^2}{25}=1\\ \frac{{\left(y+5\right)}^2}{25}=\frac{{\left(x-2\right)}^2}{4}-1\\ \frac{{\left(y+5\right)}^2}{25}=\frac{{\left(x-2\right)}^2-4}{4}\\ \left(y+5\right)=\pm 5\sqrt{\frac{{\left(x-2\right)}^2-4}{4}}\end{array}$

Simplify further to find the value of y.

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

Use the above equation and the asymptotes to draw the graph of the hyperbola.

  

Figure (1)

Therefore The center of the hyperbola is $\left(2,-5\right)$ , the vertices of the hyperbola is $\left(4,-5\right)$ , $\left(0,-5\right)$ , the foci of the hyperbola is $\left(2+\sqrt{29},-5\right)$ and $\left(2-\sqrt{29},-5\right)$ , the asymptotes of the hyperbola is $y=\frac{5x}{2}-10$ , $y=-\frac{5x}{2}$ and the graph of the hyperbola is shown in figure (1).

Verify the result using graphical utility:

Draw the graph for the hyperbola using the graphical utility.

  

From the above, it is clear that both the graphs are similar.

Hence, proved.