Chapter 8.3, Problem 68E

Solution

To calculate the slopes of given equation ${\frac{x}{4}}^2-{\frac{y}{3}}^2=1$ .

The slopes is $c=\pm \frac{\sqrt{3}}{2}$ .

Given:

The given equation is:

  ${\frac{x}{4}}^2-{\frac{y}{3}}^2=1$

Concept Used:

The general form of hyperbola asymptote is:

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

So, after comparing, we get.

  $\frac{x^2}{2^2}-\frac{y^2}{{\left(\sqrt{3}\right)}^2}=1$

The vertices in given general form can be written as:

Vertices $=\left(h+a,k\right),\left(h-a,k\right)$ .

Here, $\left(h,k\right)=\left(0,0\right)$ .

The values of $\left(a,b\right)=\left(2,\sqrt{3}\right)$ .

Then, the vertices will be calculated as:

  $\begin{array}{c}Vertices=\left(0+2,0\right),\left(0-2,0\right)\\ =\left(2,0\right),\left(-2,0\right)\end{array}$

The focus can be calculated as:

Foci $=\left(h+c,k\right),\left(h-c,k\right)$ .

  $c=\sqrt{a^2+b^2}$

Calculation:

As per the given problem,

  $\begin{array}{l}c=\sqrt{4+3}\\ c=\pm \sqrt{7}\end{array}$

The slope of the asymptotes of hyperbola is $m=\pm \frac{b}{a}$ .

Thus $m=+\frac{\sqrt{3}}{2},-\frac{\sqrt{3}}{2}$

Using the graphing calculator,

  

Thus, the slope of asymptotes of hyperbola $c=\pm \frac{\sqrt{3}}{2}$ .

The correct option is C.