Chapter 9.3, Problem 41E

To determine

To find:The standard form of the equation of hyperbola when the vertices are $\left(\pm 1,0\right)$ and asymptotes is $y=\pm 5x$ .

Answer to Problem 41E

The standard form of the equation of hyperbola is $\frac{x^2}{1}-\frac{y^2}{5^2}=1$ .

Explanation of Solution

Given information:

Thevertices of hyperbola are $\left(\pm 1,0\right)$ and asymptotes of hyperbola is $y=\pm 5x$

Calculation:

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

Since the given vertices are $\left(-1,0\right)$ and $\left(1,0\right)$ .So by comparing it we get,

  $a=1$ , $h=0$ and $k=0$

The asymptotes of the hyperbola is given by $y=\pm \frac{b}{a}\left(x-h\right)+k$ . The asymptotes of the hyperbola given is $y=\pm 5x$ So by comparing it with the above equation, we get

  $a=1$ , $b=5$ , $h=0$ and $k=0$ .

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

Substitute $a=1$ , $b=5$ , $h=0$ and $k=0$ to find the standard equation of the hyperbola

  $\begin{array}{c}\frac{{\left(x-0\right)}^2}{1^2}-\frac{{\left(y-0\right)}^2}{5^2}=1\\ \frac{x^2}{1}-\frac{y^2}{25}=1\end{array}$

Therefore, the standard form of the equation of hyperbola is $\frac{x^2}{1}-\frac{y^2}{5^2}=1$ .