Chapter 9.3, Problem 32E

(a)

To determine

To find: The standard equation of hyperbola $25x^2-4y^2=100$ .

Answer to Problem 32E

The standard form of the hyperbola is $\frac{x^2}{4}-\frac{y^2}{25}=1$ .

Explanation of Solution

Given information:

The equation of the hyperbola is $25x^2-4y^2=100$ .

Calculation:

Write the equation of the hyperbola.

  $\begin{array}{c}25x^2-4y^2=100\\ \frac{x^2}{\left(\frac{100}{25}\right)}-\frac{y^2}{\left(\frac{100}{4}\right)}=1\\ \frac{x^2}{4}-\frac{y^2}{25}=1\end{array}$

Therefore, the standard form of the hyperbola is $\frac{x^2}{4}-\frac{y^2}{25}=1$ .

(b)

To determine

To find: The center, vertices, foci, asymptotes of the hyperbola $25x^2-4y^2=100$ .

Answer to Problem 32E

The center of the hyperbola is $\left(0,0\right)$ , the vertices of the hyperbola is $\left(-2,0\right)$ , $\left(2,0\right)$ , the foci of the hyperbola is $\left(-\sqrt{29},0\right)$ and $\left(\sqrt{29},0\right)$ , the asymptotes of the hyperbola is $y=\frac{5x}{2}$ , $y=\frac{-5x}{2}$ .

Explanation of Solution

Given information:

Given equation of hyperbola is $25x^2-4y^2=100$ .

Calculation:

Compare the given equation with the standard equation of hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ .

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

The center of the hyperbola is given as $\left(0,0\right)$ .

The vertices of the hyperbola is given by $\left(-a,0\right)$ and $\left(a,0\right)$ .

So the vertices are $\left(-2,0\right)$ and $\left(2,0\right)$ .

The foci of the hyperbola is given by $\left(-c,0\right)$ and $\left(c,0\right)$ , where $c^2=a^2+b^2$ . Calculate the value of $c$ .

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

Hence the foci are $\left(-\sqrt{29},0\right)$ and $\left(\sqrt{29},0\right)$ .

The asymptotes of the hyperbola is given by $y=\pm \frac{b}{a}x$ .

Substitute $2$ for $a$ and $5$ for $b$ to find the asymptotes.

  $y=\pm \frac{5}{2}x$

Therefore, the center of the hyperbola is $\left(0,0\right)$ , the vertices of the hyperbola is $\left(-2,0\right)$ , $\left(2,0\right)$ , the foci of the hyperbola is $\left(-\sqrt{29},0\right)$ and $\left(\sqrt{29},0\right)$ , the asymptotes of the hyperbola is $y=\frac{5x}{2}$ , $y=\frac{-5x}{2}$ .

(c)

To determine

To draw: The graph of the hyperbola $25x^2-4y^2=100$ .

Answer to Problem 32E

The graph of the hyperbola is shown in figure (1).

Explanation of Solution

Given information:

The equation of the hyperbola is $25x^2-4y^2=100$ .

Calculation:

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

  $\begin{array}{c}25x^2-4y^2=100\\ \frac{x^2}{4}-\frac{y^2}{25}=1\\ 4y^2=25x^2-100\\ y^2=\frac{25x^2-100}{4}\end{array}$

Take square root on both the sides and simplify.

  $y=\pm \frac{\sqrt{25x^2-100}}{2}$

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

  

Figure (1)

Therefore, 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.