Chapter 11.3, Problem 4E

Label the vertices, foci, and asymptotes on the graphs given for the hyperbolas in Exercises 2 and 3.

1. (a) $\frac{x^2}{4^2}-\frac{y^2}{3^2}=1$

1. (b) $\frac{y^2}{4^2}-\frac{x^2}{3^2}=1$

To determine

To label: The vertices, foci and asymptotes on the graph for the hyperbola $\frac{x^2}{4^2}-\frac{y^2}{3^2}=1$.

Explanation of Solution

Definition used:

“The equation of the hyperbola with center at the origin $\left(0,0\right)$, vertices $(-a,0)$ and $\left(a,0\right)$, foci $F_1\left(-c,0\right)$ and $F_2\left(c,0\right)$ is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, where $c=\sqrt{a^2+b^2}$, $a>0$, $b>0$.”

Note that, the asymptotes of the ellipse $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ are $y=-\frac{b}{a}x$ and $y=\frac{b}{a}x$.

Calculation:

Compare the equation $\frac{x^2}{4^2}-\frac{y^2}{3^2}=1$ with $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$,

$\begin{array}{l}{a}^2=4^2\\ a=\pm 4\end{array}$

Therefore, by the definition stated above the hyperbola $\frac{y^2}{4^2}-\frac{x^2}{3^2}=1$ has the vertices $\left(-4,0\right)$ and $\left(4,0\right)$. The value of $b^2=3^2$.

By the definition of hyperbola,

$\begin{array}{c}c=\sqrt{4^2+3^2}\\ =\sqrt{16+9}\\ =\sqrt{25}\\ =\pm 5\end{array}$

Hence, the foci are $\left(-5,0\right)$ and $\left(5,0\right)$.

Therefore, the vertices and the foci of the hyperbola $\frac{y^2}{4^2}-\frac{x^2}{3^2}=1$ are $\left(-4,0\right)$, $\left(4,0\right)$ and $\left(-5,0\right)$ and $\left(5,0\right)$ respectively.

By the note stated above, the equation of asymptotes of the ellipses are, $y=-\frac{b}{a}x$ and $y=\frac{b}{a}x$ with $a=4$, $b=3$.

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

By writing the above equations in standard form the equation of asymptotes becomes, $3x+4y=0$ and $3x-4y=0$.

The graph of the hyperbola $\frac{x^2}{4^2}-\frac{y^2}{3^2}=1$ is drawn by graphic calculator as shown in Figure 1 below.

From Figure 1, the vertices $\left(-4,0\right)$ and $\left(4,0\right)$, the foci are $\left(-5,0\right)$ and $\left(5,0\right)$ and the asymptotes are $3x+4y=0$ and $3x-4y=0$.

To determine

To label: The vertices, foci and asymptotes on the graph for the hyperbola $\frac{y^2}{4^2}-\frac{x^2}{3^2}=1$.

Explanation of Solution

Definition used:

“The equation of the hyperbola with center at the $\left(0,0\right)$ vertices $\left(0,-a\right)$ and $\left(0,a\right)$ and foci $F_1\left(0,-c\right)$ and $F_2\left(0,c\right)$ is $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$, where $c=\sqrt{a^2+b^2}$, $a>0$, $b>0$.”

Note that, the asymptotes of the ellipse $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ are $y=-\frac{a}{b}x$ and $y=\frac{a}{b}x$.

Calculation:

Compare the equation $\frac{y^2}{4^2}-\frac{x^2}{3^2}=1$ with $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$.

$\begin{array}{l}{a}^2=4^2\\ a=\pm 4\end{array}$

Therefore, by the definition stated above the hyperbola $\frac{y^2}{4^2}-\frac{x^2}{3^2}=1$ has the vertices $\left(0,-4\right)$ and $\left(0,4\right)$. The value of $b^2=3^2$.

By the definition of hyperbola,

$\begin{array}{l}c=\sqrt{4^2+3^2}\\ c=\sqrt{16+9}\\ c=\sqrt{25}\\ c=\pm 5\end{array}$

Hence, the foci are $\left(0,-5\right)$ and $\left(0,5\right)$.

Therefore, the vertices and the foci of the hyperbola $\frac{y^2}{4^2}-\frac{x^2}{3^2}=1$ are $\left(0,-4\right)$ and $\left(0,4\right)$ and $\left(0,-5\right)$ and $\left(0,5\right)$ respectively.

By the note stated above, the equation of asymptotes of the ellipse are, $y=-\frac{a}{b}x$ and $y=\frac{a}{b}x$ with $a=4$, $b=3$.

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

By writing the above equations in standard form the equation of asymptotes becomes,

$4x+3y=0$ and $4x-3y=0$.

The graph of the ellipse $\frac{y^2}{4^2}-\frac{x^2}{3^2}=1$ is drawn by graphic calculator as shown in Figure 2 below.

Since the y-axis is the transverse axis, the point of intersection of the hyperbola with y- axis $(0,-4)$ and $(0,4)$ are marked as vertices and $(0,-5)$, $(0,5)$ are marked as foci.

Also, the lines $4x+3y=0$ and $4x-3y=0$ are drawn as asymptotes on the above Figure 2.