Chapter 10.3, Problem 8GCE

To determine

Give the explanation for the given statement.

Explanation of Solution

Given:

The given equation is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and the given condition is that the interchanging of $a$ and $b$ .

Calculation:

Original equation is

  $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Interchange $a$ with $b$ .

New equation is

  $\frac{y^2}{a^2}+\frac{x^2}{b^2}=1$

If $a^2$ is the denominator of the $x$ term, the major axis is parallel to the x-axis. If $a^2$ is the denominator of the $y$ term, the major axis is parallel to the y-axis.

Example

Original equation is

  $\frac{x^2}{32}+\frac{y^2}{18}=1$

In the equation, $a^2$ is the denominator of the $x$ term.

  $a^2=32$ , $b^2=18$ and $c^2=a^2-b^2=\sqrt{{32}^2-{18}^2}=10\sqrt{7}$

Center $\to \left(h,k\right)=\left(0,0\right)$

Major axis vertices $\to \left(h\pm a,k\right)=\left(0\pm 4\sqrt{2},0\right)=\left(\pm 4\sqrt{2},0\right)$

Minor axis vertices $\to \left(h,k\pm b\right)=\left(0,0\pm 3\sqrt{2}\right)=\left(0,\pm 3\sqrt{2}\right)$

Graph of the equation is given below.

  

Original equation is

  $\frac{x^2}{18}+\frac{y^2}{32}=1$

In the equation, $a^2$ is the denominator of the $y$ term.

  $a^2=32$ , $b^2=18$ and $c^2=a^2-b^2=\sqrt{{32}^2-{18}^2}=10\sqrt{7}$

Center $\to \left(h,k\right)=\left(0,0\right)$

Major axis vertices $\to \left(h,k\pm a\right)=\left(0,0\pm 4\sqrt{2}\right)=\left(0,\pm 4\sqrt{2}\right)$

Minor axis vertices $\to \left(h\pm b,k\right)=\left(0\pm 3\sqrt{2},0\right)=\left(\pm 3\sqrt{2},0\right)$

Graph of the equation are given below.