Chapter H, Problem 1PS

To determine

Draw a graph of the hyperbola using given properties.

Answer to Problem 1PS

Theequation of the hyperbola is $\frac{x^2}{16}-\frac{y^2}{9}=1$ .

Explanation of Solution

Given:

The distance between two foci is $10$ units and the curve with the difference of the distances from the two foci $8$ units.

Calculation:

A hyperbola is the set of all points in a plane such that, for each point on the hyperbola, the difference of its distances from two fixed points is a constant.

From definition,

If $\left(x,y\right)$ is any point on the curve and the two fixed points (foci) are $\left(\mp c,0\right)$ , then the difference of $\left(x,y\right)$ from the two fixed points $\left(\mp c,0\right)$ is constant $\left(2a\right)$ .

  $\begin{array}{l}\sqrt{{\left(x+c\right)}^2+{\left(y-0\right)}^2}-\sqrt{{\left(x-c\right)}^2+{\left(y-0\right)}^2}=2a\\ \Rightarrow \frac{x^2}{a^2}-\frac{y^2}{c^2-a^2}=1\left[b^2=c^2-a^2\right]\end{array}$

Now,

The distance between two foci $=2c$

From question,

  $\begin{array}{l}2c=10\\ \Rightarrow \frac{2c}{2}=\frac{10}{2}\\ \Rightarrow c=5\end{array}$

And

  $\begin{array}{l}2a=8\\ \Rightarrow \frac{2a}{2}=\frac{8}{2}\\ \Rightarrow a=4\end{array}$

  $\begin{array}{l}{b}^2=c^2-a^2\\ b^2=5^2-4^2\\ \Rightarrow b^2=25-16\\ \Rightarrow b^2=9\end{array}$

The equation of hyperbola is

  $\begin{array}{l}\frac{x^2}{a^2}-\frac{y^2}{c^2-a^2}=1\\ \Rightarrow \frac{x^2}{a^2}-\frac{y^2}{b^2}=1\\ \Rightarrow \frac{x^2}{16}-\frac{y^2}{9}=1\end{array}$

The standard form of a hyperbola is

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

  $\left(h,k\right)$ is the center of the hyperbola.

Rewrite the given equation.

  $\begin{array}{l}\frac{{\left(x-0\right)}^2}{16}-\frac{{\left(y-0\right)}^2}{9}=1\\ \Rightarrow \frac{{\left(x-0\right)}^2}{{\left(4\right)}^2}-\frac{{\left(y-0\right)}^2}{{\left(3\right)}^2}=1\end{array}$

  $\begin{array}{l}a=4\\ b=3\\ h=0\\ k=0\end{array}$

Center of the hyperbola $\left(h,k\right)=\left(0,0\right)$ .

The distance from the center to a focus.

  $c=5$

Vertex $v_1=\left(h+a,k\right)=\left(0+4,0\right)=\left(4,0\right)\text{and}{v}_2=\left(h-a,k\right)=\left(0-4,0\right)=\left(-4,0\right)$

Focus $f_1=\left(\mp 5,0\right)$

Eccentricity

  $e=\frac{c}{a}=\frac{5}{4}$

Asymptotes $y=\pm \frac{b\left(x-h\right)}{a}+k=\pm \frac{3\left(x-0\right)}{4}+0=\pm \frac{3}{4}x$

  $\begin{array}{l}y=\frac{3}{4}x\\ y=-\frac{3}{4}x\end{array}$

The graph of the curve is given below.

  

Hence the equation of the hyperbola is $\frac{x^2}{16}-\frac{y^2}{9}=1$ .