Chapter 11, Problem 30CCSR

Solution

To calculate: The vertices, intercepts, asymptotes, and foci of the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ .

The vertices, intercepts, asymptotes, and foci of the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ are

  $\begin{array}{l}\text{Vertices }(4,0)(-4,0)\\ \text{Foci }(5,0)(-5,0)\\ \text{Intercept }(4,0)(-4,0)\\ \text{Asymptotes }y=\pm \frac{3}{4}x\end{array}$

Given:The equation of the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ .

Concept used:

The hyperbola with equation $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ has

Foci at $(-ae,0)(ae,0)$ where $e=\frac{\sqrt{a^2+b^2}}{a}$

Vertices and intercepts at $(-a,0)(a,0)$

Asymptotes at $y=\pm \frac{b}{a}x$

Calculation:

Given hyperbola equation $\frac{x^2}{16}-\frac{y^2}{9}=1$ is in the form of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ where,

  $a=4\text{ and }b=3$

The eccentricity of hyperbola e is calculated as shown below

  $\begin{array}{l}e=\frac{\sqrt{a^2+b^2}}{a}\\ e=\frac{\sqrt{16+9}}{4}=\frac{5}{4}\end{array}$

The foci of the hyperbola are

  $\begin{array}{l}(-ae,0)\text{ =}\left(-4\times \frac{5}{4},0\right)=\left(-5,0\right)\\ (ae,0)\text{=}\left(4\times \frac{5}{4},0\right)=\left(5,0\right)\end{array}$

The vertices and the intercepts of the hyperbola are

  $\begin{array}{l}(-a,0)=(-4,0)\\ (a,0)=(4,0)\end{array}$

The asymptotes of the hyperbola are

  $\begin{array}{l}y=\pm \frac{b}{a}x\\ y=\pm \frac{3}{4}x\end{array}$

Conclusion:Using the formulas and the given equation of hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ has following

  $\begin{array}{l}\text{Vertices }(4,0)(-4,0)\\ \text{Foci }(5,0)(-5,0)\\ \text{Intercept }(4,0)(-4,0)\\ \text{Asymptotes }y=\pm \frac{3}{4}x\end{array}$