Chapter 9.4, Problem 46HP

To determine

To find: the explanation of the domain and range for an ellipse with equation.

Answer to Problem 46HP

The domain and range for an ellipse with equation are $\left\{x|-a\le x\le a\right\}$ and $\left\{y|-b\le y\le b\right\}$

Explanation of Solution

Given information:

The given equation for an ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Concept used:

Write the standard form of the equation of the ellipse whose orientation is along the horizontal axis is,

  $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$

The coordinate of the foci is $\left(h\pm c,k\right)$

The value of $c$ is calculated as,

  $c=\sqrt{a^2-b^2}$

The vertices are at $\left(h\pm a,k\right)$ and the co-vertices are at $\left(h,k\pm b\right)$ .

Write the standard form of the equation of the ellipse whose orientation is along the vertical axis is,

  $\frac{{\left(y-k\right)}^2}{a^2}+\frac{{\left(x-h\right)}^2}{b^2}=1$

The coordinate of the foci is $\left(h,k\pm c\right)$

The vertices are at $\left(h,k\pm a\right)$ and the co-vertices are at $\left(h\pm b,k\right)$ .

The value of $c$ is calculated as,

  $c=\sqrt{a^2-b^2}$

The given equation of an ellipse is,

  $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\mathrm{......}\left(1.\right)$

The equation as a function of $x$ is calculated as,

  $\begin{array}{l}1-\frac{x^2}{a^2}=\frac{y^2}{b^2}\\ y^2=b^2\left(\frac{a^2-x^2}{a^2}\right)\\ y=\pm \frac{b}{a}\sqrt{\left(a^2-x^2\right)}\end{array}$

  $y$ as a function of $x$ is

  $y=\pm \frac{b}{a}\sqrt{\left(a^2-x^2\right)}\mathrm{.....}\left(2.\right)$

The expression in the radical is non-negative,

  $\begin{array}{l}{a}^2-x^2\ge 0\\ a^2\ge x^2\\ -a\le x\le a\end{array}$

The domain of an ellipse is $\left\{x|-a\le x\le a\right\}$

The zero is the minimum value of the radical,

To calculated the value of $y$ to put the minimum value of radical $\left(0\right)$ in the function (2.)

  $\begin{array}{l}y=\pm \frac{b}{a}\sqrt{\left(a^2-x^2\right)}\\ y=\pm \frac{b}{a}\times 0\\ y=0\end{array}$

To calculate the value of $y$ by putting the maximum value of radical $\left(a\right)$ in the equation.(2.)

  $\begin{array}{l}y=\pm \frac{b}{a}\sqrt{\left(a^2-x^2\right)}\\ y=\pm \frac{b}{a}\times a\\ y=\pm b\end{array}$

The function $y$ lies at between $-b$ and $b$

The range for an ellipse is $\left\{y|-b\le y\le b\right\}$

Thus, the domain and range for an ellipse with equation are $\left\{x|-a\le x\le a\right\}$ and $\left\{y|-b\le y\le b\right\}$