Chapter 9.5, Problem 3OE

To determine

To calculate: To find the x- and y-intercepts of the hyperbola and the equation of its asymptotes and to tell on which of the coordinate axes its foci lie

Answer to Problem 3OE

x-intercept is $1$ and y-intercept is $5$ and foci lies on y-axis and equation of asymptotes are $y=\pm \frac{1}{5}x$

Explanation of Solution

Given information: Equation of hyperbola is $x^2-25y^2+25=0$

Formula Used:

The hyperbola with center $\left(0,0\right)$ and foci $\left(-c,0\right)$ and $\left(c,0\right)$ and difference of focal radii $2a$ has the equation:

  $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ , where $b^2=c^2-a^2$

Equation of the asymptotes are $y=\frac{b}{a}x$ and $y=-\frac{b}{a}x$

The hyperbola with center $\left(0,0\right)$ and foci $\left(0,-c\right)$ and $\left(0,c\right)$ and difference of focal radii $2a$ has the equation:

  $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ , where $b^2=c^2-a^2$

Equation of the asymptotes are $y=\frac{a}{b}x$ and $y=-\frac{a}{b}x$

Calculation:

Equation of hyperbola is given as follows:

  $x^2-25y^2+25=0$

Divide both sides by $25$ :

  $\begin{array}{l}\frac{x^2}{25}-\frac{y^2}{1}+1=0\\ \frac{y^2}{1}-\frac{x^2}{25}=1\\ \frac{y^2}{1^2}-\frac{x^2}{5^2}=1\end{array}$

Comparing the equation with standard form of the equation of hyperbola $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ :

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

Foci is calculated as follows:

  $\begin{array}{l}{b}^2=c^2-a^2\\ c^2=a^2+b^2\\ c^2=1^2+5^2\\ c^2=1+25\\ c^2=26\\ c=\pm \sqrt{26}\end{array}$

Thus, Foci are $\left(0,\sqrt{26}\right)$ and $\left(0,-\sqrt{26}\right)$ and foci lies on y-axis

Also, asymptotes are calculated as follows:

  $y=\frac{a}{b}x$ and $y=-\frac{a}{b}x$

Substituting the values:

  $y=\frac{1}{5}x$

and

  $y=-\frac{1}{5}x$

Conclusion:

Hence, x-intercept is $1$ and y-intercept is $5$ and foci lies on y-axis and equation of asymptotes are $y=\pm \frac{1}{5}x$