Chapter 9.2, Problem 46E

To determine

To find the center, vertices, foci and eccentricity of the ellipse and graph the ellipse.

Answer to Problem 46E

The center of the ellipse is $\left(-\frac{2}{3},2\right)$ .

The vertices are $\left(-\frac{2}{3},3\right)\text{ and }\left(-\frac{2}{3},1\right)$ .

The foci are $\left(-\frac{2}{3},2+\frac{\sqrt{3}}{2}\right)\text{ and }\left(-\frac{2}{3},2-\frac{\sqrt{3}}{2}\right)$ .

The eccentricity is $\frac{c}{a}=\frac{\sqrt{3}}{2}$ .

Explanation of Solution

Given information:

The equation of the ellipse is:

  $36x^2+9y^2+48x-36y+43=0$

Formula used:

The following formula is used:

  $\text{Eccentricity,}e=\frac{c}{a}$

Calculation:

Write the equation in standard form by completing the square.

  $\begin{array}{l}36x^2+9y^2+48x-36y+43=0\\ 36\left(x^2+\frac{4}{3}x\right)+9\left(y^2-4y\right)=-43\\ 36\left(x^2+\frac{4}{3}x+\frac{4}{9}\right)+9\left(y^2-4y+4\right)=-43+36\left(\frac{4}{9}\right)+9\left(4\right)\\ 36{\left(x+\frac{2}{3}\right)}^2+9{\left(y-2\right)}^2=9\\ \frac{36{\left(x+\frac{2}{3}\right)}^2}{9}+\frac{9{\left(y-2\right)}^2}{9}=1\\ \frac{{\left(x+\frac{2}{3}\right)}^2}{\frac{1}{4}}+\frac{{\left(y-2\right)}^2}{1}=1\end{array}$

The ellipse has a major vertical axis, where

  $h=-\frac{2}{3},k=2,a=1,b=\frac{1}{2}\text{ and }c=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{14}}{2}$ .

The center of the ellipse is $\left(-\frac{2}{3},2\right)$ .

The vertices are $\left(-\frac{2}{3},3\right)\text{ and }\left(-\frac{2}{3},1\right)$ .

The foci are $\left(-\frac{2}{3},2+\frac{\sqrt{3}}{2}\right)\text{ and }\left(-\frac{2}{3},2-\frac{\sqrt{3}}{2}\right)$ .

The eccentricity is $\frac{c}{a}=\frac{\sqrt{3}}{2}$ .

To graph the ellipse solve for $y$ :

  $\begin{array}{l}4{\left(x+\frac{2}{3}\right)}^2+{\left(y-2\right)}^2=1\\ {\left(y-2\right)}^2=1-4{\left(x+\frac{2}{3}\right)}^2\\ y-2=\pm \sqrt{1-4{\left(x+\frac{2}{3}\right)}^2}\\ y=\sqrt{1-4{\left(x+\frac{2}{3}\right)}^2}+2\text{ or }-\sqrt{1-4{\left(x+\frac{2}{3}\right)}^2}+2\end{array}$

The graph is given below: