Chapter 11.3, Problem 3E

To determine

To fill: The graph of the equation $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ with $a>0$, $b>0$ is hyperbola with _____ (horizontal / vertical ) transverse axis, vertices ___ and ___ and foci ___, where ___. So the graph of $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ is a hyperbola with vertices ___ and ____ and foci ___ and ___.

Answer to Problem 3E

The graph of the equation $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ with $a>0$, $b>0$ is hyperbola with vertical (horizontal / vertical ) transverse axis, vertices $\left(0,-a\right)$ and $\left(0,a\right)$ and foci $\left(0,\pm c\right)$, where $c=\sqrt{a^2+b^2}$. So the graph of $\frac{y^2}{4^2}-\frac{x^2}{3^2}=1$ is a hyperbola with vertices $\left(0,-4\right)$ and $\left(0,4\right)$ and foci $\left(0,-5\right)$ and $\left(0,5\right)$.

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 $c=\sqrt{a^2+b^2}$, $a>0$, $b>0$.”

Calculation:

The given equation of the hyperbola is $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ (1)

To get the intercept on the y- axis, put $x=0$ in equation .

Now equation (1) becomes,

$\begin{array}{c}\frac{y^2}{a^2}=1\\ y^2=a^2\\ y=\pm a\end{array}$

Therefore, equation (1) intersects y- axis at $\left(0,-a\right)$ and $\left(0,a\right)$.

Similarly, to get the intercept on the x- axis, put $y=0$ in equation .

Now equation (1) becomes,

$\begin{array}{c}-\frac{x^2}{b^2}=1\\ x^2=-b^2\\ x=\pm \sqrt{-b^2}\end{array}$

Since the square root of negative number becomes imaginary there is no real roots.

Hence the equation (1) do not have the intercept on x-axis.

Now, the equation (1) has the vertices $\left(0,-a\right)$ and $\left(0,a\right)$, and vertical (y-axis) transverse axis.

Therefore, the graph of the equation $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ with $a>0$, $b>0$ is a hyperbola with vertical (horizontal / vertical ) transverse axis, vertices $\left(0,-a\right)$ and $\left(0,a\right)$ and foci $\left(\pm c,0\right)$ also by the definition stated above $c=\sqrt{a^2+b^2}$.

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, the value of $c$ computed as follows.

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

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.