Chapter 9, Problem 29RE

To determine

To find: The centre, vertices, foci, and eccentricity of the ellipse $\frac{{\left(x-3\right)}^2}{36}+\frac{{\left(y-2\right)}^2}{4}=1$ , and sketch it graph.

Answer to Problem 29RE

Center: $\left(3,2\right)$ .

Vertices: $\left(-3,2\right)$ and $\left(9,2\right)$ ,

Foci: $\left(3-4\sqrt{2},2\right)$ and $\left(3+4\sqrt{2},2\right)$

Eccentricity: $\frac{2\sqrt{2}}{3}$

Explanation of Solution

Given:

  $\frac{{\left(x-3\right)}^2}{36}+\frac{{\left(y-2\right)}^2}{4}=1$

Calculation:

The standard equation of an ellipse is given by $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$ .

The center of this ellipse is $\left(h,k\right)$ .

Write the given ellipse in standard form.

  $\frac{{\left(x-3\right)}^2}{6^2}+\frac{{\left(y-2\right)}^2}{2^2}=1$

Compare the given ellipse $\frac{{\left(x-3\right)}^2}{6^2}+\frac{{\left(y-2\right)}^2}{2^2}=1$ with the standard form $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$ . This gives,

  $h=3$ , $k=2$ , $a=6$ and $b=2$ .

So, the center of the ellipse is $\left(3,2\right)$ .

So, the vertices of the ellipse are $\left(3-6,2\right)=\left(-3,2\right)$ and $\left(3+6,2\right)=\left(9,2\right)$ .

Now, find the value of $c$ using the equation $b^2-c^2=a^2$ .

  $\begin{array}{l}{a}^2-c^2=b^2\\ 6^2-c^2={\left(2\right)}^2\\ \Rightarrow c=\sqrt{36-4}=\sqrt{32}=4\sqrt{2}\end{array}$

So, the foci of the ellipse are $\left(3-4\sqrt{2},2\right)$ and $\left(3+4\sqrt{2},2\right)$ .

Then, the value of the eccentricity is $e=\frac{c}{a}=\frac{4\sqrt{2}}{6}=\frac{2\sqrt{2}}{3}$ .

Now, graph the ellipse as shown below.

  

Now, label the center, vertices and foci on the graph to verify the result as shown below.