Chapter 11.3, Problem 25E

To determine

To find: the equation for the hyperbola.

Answer to Problem 25E

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

Explanation of Solution

Given information:

From the figure, provided in the question it is observed that

Vertices of hyperbola : $\left(\pm 5,0\right)$

asymptotes of hyperbola : $y=\pm \frac{1}{2}x$

Calculation:

since from the given vertices it is observed that the vertices of the hyperbola are on the x-axis, so the hyperbola has a horizontal transverse axis.

So the equation is of the form,

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

For transverse axis hyperbola, the asymptote

  $y=\pm \frac{b}{a}x$

So from the $y=\pm \frac{1}{2}x$ asymptote it can be observed that ,

  $\frac{b}{a}=\frac{1}{2}$

Since $a=5$ , now to find $b$ substitute $a=5$ ,

  $\begin{array}{l}\frac{b}{5}=\frac{1}{2}\\ b=\frac{5}{2}\end{array}$

so the equation of hyperbola is,

  $\begin{array}{l}\frac{x^2}{{\left(a\right)}^2}-\frac{y^2}{b^2}=1\\ \frac{x^2}{{\left(5\right)}^2}-\frac{y^2}{{\left(\frac{5}{2}\right)}^2}=1\text{ [substitute]}\\ \frac{x^2}{25}-\frac{y^2}{\frac{25}{4}}=1\\ \frac{x^2}{25}-\frac{4y^2}{25}=1\text{ [simplify]}\end{array}$

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