Chapter 9, Problem 27RE

To determine

To find: The centre, vertices, foci, and eccentricity of the ellipse $\frac{x^2}{25}+\frac{y^2}{36}=1$ , and sketch it graph.

Answer to Problem 27RE

Center: $\left(0,0\right)$ .

Vertices: $\left(0,-6\right)$ and $\left(0,6\right)$ ,

Foci: $\left(0,-\sqrt{11}\right)$ and $\left(0,\sqrt{11}\right)$

Eccentricity: $\frac{\sqrt{11}}{6}$

Explanation of Solution

Given:

  $\frac{x^2}{25}+\frac{y^2}{36}=1$

Calculation:

Write the given ellipse in standard form.

  $\frac{{\left(x-0\right)}^2}{5^2}+\frac{{\left(y-0\right)}^2}{6^2}=1$

Compare the given ellipse $\frac{x^2}{25}+\frac{y^2}{36}=1$ with the standard form $\frac{{\left(x-h\right)}^2}{b^2}+\frac{{\left(y-k\right)}^2}{a^2}=1$ . This gives,

  $h=0$ , $k=0$

  $a=6$ and $b=5$ .

So, the center of the ellipse is $\left(0,0\right)$ .

Since $b>a$ , the ellipse is vertical. So, the vertices of the ellipse are $\left(0,-6\right)$ and $\left(0,6\right)$ .

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

  $\begin{array}{l}{a}^2-c^2=b^2\\ 6^2-c^2=5^2\\ \Rightarrow c=\sqrt{36-25}=\sqrt{11}\end{array}$

So, the foci of the ellipse are $\left(0,-\sqrt{11}\right)$ and $\left(0,\sqrt{11}\right)$ .

Then, the value of the eccentricity is $e=\frac{c}{a}=\frac{\sqrt{11}}{6}$ .

Now, graph the ellipse as shown below.

  

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