Chapter 14.2, Problem 72SR

To determine

To find :Coordinates of vertices, foci equation of asymptotes for the hyperbola and its graph.

Answer to Problem 72SR

  $\begin{array}{l}\text{Vertices, (6, 0) and (}-6,0\text{)}\\ \text{foci, (}\sqrt{\text{37}}\text{,}\text{0) and (}-\sqrt{37}\text{, 0)}\\ \text{Equation of asymptote, }y=\pm \frac{x}{6}\end{array}$

Explanation of Solution

Given information :

We know that, $x^2-36y^2=36$

Calculation :

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

Hyperbola is horizontal hyperbola.

Putting $\text{y}\text{ = 0}$

  $\begin{array}{l}\frac{{\text{x}}^{\text{2}}}{36}-\frac{0}{1}=1\\ \Rightarrow x^2=36\Rightarrow x=\pm 6\\ \text{Vertices (+6, 0), (}-6,0\text{)}\\ \text{For foci, }c=\pm \sqrt{a^2+b^2}\\ c=\pm \sqrt{36+1}=\pm \sqrt{37}\\ \text{foci}(\sqrt{37},0),(-\sqrt{37},0)\end{array}$

Equation of asymptotes are

  $y=\pm \frac{1}{6}x$