Chapter 8.5, Problem 33E

Solution

(a)

To explain : The effects of the indicated change on the shape of the hyperbola and on the locations of the vertices and foci.

When we change $36$ to $4$ , location of the vertices remain the same, whereas the foci has moved from $\left(\pm 9,0\right)\text{ to }\left(\pm \sqrt{13},0\right)$ . As a result, the hyperbola becomes narrower.

Given information :

The equation of hyperbola: $\frac{x^2}{9}-\frac{y^2}{36}=1$ ; change $36$ to $4$ .

Explanation :

Given equation is $\frac{x^2}{9}-\frac{y^2}{36}=1$ .

Rewrite the given equation in standard form.

  $\frac{x^2}{3^2}-\frac{y^2}{6^2}=1$

The equation is now of the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ . An equation of this form has vertices at $\left(\pm a,0\right)$ .

Determine the value of $a$ and $b$ by comparing $\frac{x^2}{3^2}-\frac{y^2}{6^2}=1$ with the standard form.

  $\begin{array}{l}{a}^2=3^2\\ b^2=6^2\end{array}$

The value of $a$ is $3$ and $b$ is $6$ . Thus, the vertices of the hyperbola are $\left(3,0\right)\text{ and }\left(-3,0\right)$ .

In order to find the foci, substitute $9$ for $a^2$ and $36$ for $b^2$ in the formula

  $\begin{array}{c}{c}^2=a^2+b^2\\ =9+36\\ =45\end{array}$

The foci are at $\left(\pm 9,0\right)$ .

Replace $36$ with $4$ in the given equation and rewrite the equation in the standard form.

  $\frac{x^2}{3^2}-\frac{y^2}{2^2}=1$

Compare the equation $\frac{x^2}{3^2}-\frac{y^2}{2^2}=1$ with the standard form, and find the values of $a\text{ and }b$ .

  $\begin{array}{l}{a}^2=3^2\\ b^2=2^2\end{array}$

The value of $a$ is $3$ and $b$ is $2$ . Thus, the vertices of the hyperbola are $\left(3,0\right)\text{ and }\left(-3,0\right)$ .

In order to find the foci, substitute $9$ for $a^2$ and $4$ for $b^2$ in the formula

  $\begin{array}{c}{c}^2=a^2+b^2\\ =9+4\\ =13\end{array}$

The foci are at $\left(\pm \sqrt{13},0\right)$ .

Thus, when we change $36$ to $4$ , location of the vertices remain the same, whereas the foci has moved from $\left(\pm 9,0\right)\text{ to }\left(\pm \sqrt{13},0\right)$ . As a result, the hyperbola becomes narrower.

(b)

To explain : The effects of the indicated change on the shape of the hyperbola and on the locations of the vertices and foci.

When we change $4$ to $25$ , location of the vertices remain the same, whereas the foci has moved from $\left(0,\pm \sqrt{20}\right)\text{ to }\left(0,\pm \sqrt{41}\right)$ . As a result, the hyperbola becomes wider.

Given information :

The equation of hyperbola: $\frac{y^2}{16}-\frac{x^2}{4}=1$ ; change $4$ to $25$ .

Explanation :

Given equation is $\frac{y^2}{16}-\frac{x^2}{4}=1$ .

Rewrite the given equation in standard form.

  $\frac{x^2}{4^2}-\frac{y^2}{2^2}=1$

The equation is now of the form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ . An equation of this form has vertices at $\left(0,\pm a\right)$ .

Determine the value of $a$ and $b$ by comparing $\frac{x^2}{4^2}-\frac{y^2}{2^2}=1$ with the standard form.

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

The value of $a$ is $4$ and $b$ is $2$ .

In order to find the foci, substitute $16$ for $a^2$ and $4$ for $b^2$ in the formula

  $\begin{array}{c}{c}^2=a^2+b^2\\ =16+4\\ =20\end{array}$

Thus, the vertices of the hyperbola are $\left(0,4\right)\text{ and }\left(0,-4\right)$ and the foci are at $\left(0,\pm \sqrt{20}\right)$ .

Replace $4$ with $25$ in the given equation and rewrite the equation in the standard form.

  $\frac{x^2}{4^2}-\frac{y^2}{2^2}=1$

Compare the equation $\frac{x^2}{4^2}-\frac{y^2}{2^2}=1$ with the standard form, and find the values of $a\text{ and }b$ .

  $\begin{array}{l}{a}^2=4^2\\ b^2=5^2\end{array}$

The value of $a$ is $4$ and $b$ is $5$ .

In order to find the foci, substitute $16$ for $a^2$ and $25$ for $b^2$ in the formula

  $\begin{array}{c}{c}^2=a^2+b^2\\ =16+25\\ =41\end{array}$

Thus, the vertices of the hyperbola are $\left(0,4\right)\text{ and }\left(0,-4\right)$ and the foci are at $\left(0,\pm \sqrt{41}\right)$ .

Thus, when we change $4$ to $25$ , location of the vertices remain the same, whereas the foci has moved from $\left(0,\pm \sqrt{20}\right)\text{ to }\left(0,\pm \sqrt{41}\right)$ . As a result, the hyperbola becomes wider.