Chapter 11.4, Problem 4E

To determine

To label: The vertices, foci and asymptotes on the $\frac{x^2}{4^2}-\frac{y^2}{3^2}=1$ and   $\frac{{\left(x-3\right)}^2}{4^2}-\frac{{\left(y-1\right)}^2}{3^2}=1$

Answer to Problem 4E

For the hyperbola $\frac{x^2}{4^2}-\frac{y^2}{3^2}=1$ with the center $C\left(0,0\right)$, the vertices are $V_1\left(-4,0\right)$ and $V_2\left(4,0\right)$, the foci are $F_1\left(-3,0\right)$ and $F_2\left(3,0\right)$, and the equations of asymptote are $3x+4y=0$  and $3x-4y=0$.

For the hyperbola $\frac{{\left(x-3\right)}^2}{4^2}-\frac{{\left(y-1\right)}^2}{3^2}=1$ with the center $C\left(3,1\right)$, the vertices are $V_1\left(-1,1\right)$ and $V_2\left(7,1\right)$, the foci are $F_1\left(-2,1\right)$ and $F_2\left(8,1\right)$, and the equations of asymptotes are $3x+4y-13=0$  and $3x-4y-5=0$ respectively.

Explanation of Solution

Definition used:

Definition 1:

The equation of the shifted hyperbola with center as $\left(h,k\right)$ and horizontal axis (x-axis) axis as transverse axis is $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$ , when $a>0$, $b>0$, $c^2=a^2+b^2$.

Definition 2:

The equation of the hyperbola with center at the origin, vertices $(-a,0)\text{ and }\left(a,0\right)$, and foci $F_1(-c,0)\text{ and }{F}_2(c,0)$, is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, where $c^2=a^2+b^2$, $a>0$, $b>0$.

Definition 3:

The equation of the asymptotes of the hyperbola $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$ are $\frac{\left(x-h\right)}{a}+\frac{\left(y-k\right)}{b}=0$ and $\frac{\left(x-h\right)}{a}-\frac{\left(y-k\right)}{b}=0$.

Definition 4:

The equation of the asymptotes of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ are $\frac{x}{a}+\frac{y}{b}=0$ and $\frac{x}{a}-\frac{y}{b}=0$.

Definition 5:

If $h$ and $k$ are any two positive real numbers and a graph of any equation is in terms of $x$ and $y$, then

(a) When replace $x$ by $x-h$, the graph shifts right by $h$ units.

(b) When replace $x$ by $x+h$, the graph shifts left by $h$ units.

(c) When replace $y$ by $y-k$, the graph shifts upward by $k$ units.

(d) When replace $y$ by $y+k$, the graph shifts downward by $k$ units.

Calculation:

Consider the equation of hyperbola, $\frac{x^2}{4^2}-\frac{y^2}{3^2}=1$ (1)

By the definition 2 and 4, compute the vertices, foci and asymptotes as follows.

The vertices becomes $(-4,0)\text{ and }\left(4,0\right)$.

Compare the equation (1) with the general hyperbola equation $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.

$a^2=4^2\text{ implies }a=\pm 4\text{ and }{b}^2=3^2\text{ implies b=}\pm \text{3}$.

Use $c^2=a^2+b^2$, to find the value of $c$.

$\begin{array}{l}{c}^2=16+9\\ c^2=25\\ c=5\end{array}$

By the definition 4, the equations of asymptotes of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ are $\frac{x}{a}+\frac{y}{b}=0$ and $\frac{x}{a}-\frac{y}{b}=0$.

Then, for the hyperbola $\frac{x^2}{4^2}-\frac{y^2}{3^2}=1$, the asymptotes become as follows.

$\begin{array}{l}\frac{x}{a}+\frac{y}{b}=0\\ \frac{x}{4}+\frac{y}{3}=0\\ 3x+4y=0\end{array}$

And,

$\begin{array}{l}\frac{x}{a}-\frac{y}{b}=0\\ \frac{x}{4}-\frac{y}{3}=0\\ 3x-4y=0\end{array}$

Thus, for the hyperbola $\frac{x^2}{4^2}-\frac{y^2}{3^2}=1$ with the center $C\left(0,0\right)$, the vertices are $V_1\left(-4,0\right)$ and $V_2\left(4,0\right)$, the foci are $F_1\left(-3,0\right)$ and $F_2\left(3,0\right)$, and the equations of asymptote are $3x+4y=0$  and $3x-4y=0$.

By definition 3, the equations of asymptotes of shifted hyperbola are,

$\frac{\left(x-h\right)}{a}+\frac{\left(y-k\right)}{b}=0$ and $\frac{\left(x-h\right)}{a}-\frac{\left(y-k\right)}{b}=0$.

Then, for the shifted hyperbola $\frac{{\left(x-3\right)}^2}{4^2}-\frac{{\left(y-1\right)}^2}{3^2}=1$, the asymptotes become as obtained below.

$\begin{array}{l}\frac{\left(x-h\right)}{a}+\frac{\left(y-k\right)}{b}=0\\ \frac{\left(x-3\right)}{4}+\frac{\left(y-1\right)}{3}=0(\text{Since},.a=4\text{ and }b=3)\\ 3(x-3)+4\left(y-1\right)=0\\ 3x+4y-13=0\end{array}$

And,

$\begin{array}{l}\frac{\left(x-h\right)}{a}-\frac{\left(y-k\right)}{b}=0\\ \frac{\left(x-3\right)}{4}-\frac{\left(y-1\right)}{3}=0\left(\text{Since, }a=4\text{ and }b=3\right)\\ 3(x-3)-4\left(y-1\right)=0\\ 3x-4y-5=0\end{array}$

Hence, the equations of asymptotes are $3x+4y-13=0$ and $3x-4y-5=0$.

Therefore, for the hyperbola $\frac{{\left(x-3\right)}^2}{4^2}-\frac{{\left(y-1\right)}^2}{3^2}=1$ with the center $C\left(3,1\right)$, the vertices are $V_1\left(-1,1\right)$ and $V_2\left(7,1\right)$, the foci are $F_1\left(-2,1\right)$ and $F_2\left(8,1\right)$ and the equations of asymptotes are $3x+4y-13=0$ and $3x-4y-5=0$ respectively.