Chapter 9.5, Problem 6WE

To determine

To calculate: To graph the hyperbola and show its asymptotes by dashed lines and to find the coordinates of foci

Answer to Problem 6WE

Coordinates of foci are $\left(0,\sqrt{13}\right)$ and $\left(0,-\sqrt{13}\right)$

Explanation of Solution

Given information: Equation of hyperbola is $4x^2-9y^2+36=0$

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:

  $4x^2-9y^2+36=0$

Rewriting and divide both sides of equation by $36$

  $\begin{array}{l}9y^2-4x^2=36\\ \frac{y^2}{4}-\frac{x^2}{9}=1\\ \frac{y^2}{2^2}-\frac{x^2}{3^2}=1\end{array}$

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

  $\begin{array}{l}a=2\\ b=3\end{array}$

Foci is calculated as follows:

  $\begin{array}{l}{b}^2=c^2-a^2\\ c^2=a^2+b^2\\ c^2=2^2+3^2\\ c^2=4+9\\ c^2=13\\ c=\pm \sqrt{13}\end{array}$

Thus, Foci are $\left(0,\sqrt{13}\right)$ and $\left(0,-\sqrt{13}\right)$

Also, asymptotes are calculated as follows:

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

Substituting the values:

  $y=\frac{2}{3}x$   and  $y=-\frac{2}{3}x$

Graph the hyperbola as follows:

  

Conclusion:

Hence, graph of hyperbola is sketched and coordinates of foci are $\left(0,\sqrt{13}\right)$ and $\left(0,-\sqrt{13}\right)$