Chapter 11.3, Problem 2E

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

To determine

To fill: The graph of the equation $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ with $a>0$, $b>0$ is a hyperbola with ______ (horizontal / vertical ) transverse axis, vertices ___ and ___ and foci ___, where ____. So the graph of $\frac{x^2}{4^2}-\frac{y^2}{3^2}=1$ is a hyperbola with vertices ____ and ____ and foci ____ and ____.

Answer to Problem 2E

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

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 $c=\sqrt{a^2+b^2}$, $a>0$, $b>0$.”

Calculation:

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

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

Now equation (1) becomes,

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

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

Similarly, to get the intercept on the y- axis, put $x=0$ in equation (1),

Now equation (1) becomes,

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

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

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

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

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

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.