Chapter 9, Problem 28CR

To determine

To calculate: The center, vertices, foci and eccentricity of the ellipse and sketch its graph

Answer to Problem 28CR

The center of the ellipse is $\left(-5,1\right)$ , the vertices is $\left(-5,-2\right)\text{ and }\left(-5,4\right)$ , the eccentricity is $\frac{2\sqrt{2}}{3}$ and the foci is $\left(-5,1+2\sqrt{2}\right)\text{ and }\left(-5,1-2\sqrt{2}\right)$ . The graph is represented.

Explanation of Solution

Given information:

The expression of the ellipse is $\frac{{\left(x+5\right)}^2}{1}+\frac{{\left(y-1\right)}^2}{9}=1$

Formula used:

The standard form of the equation of an ellipse is,

  $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$ [For horizontal major axis]

  $\frac{{\left(x-h\right)}^2}{b^2}+\frac{{\left(y-k\right)}^2}{a^2}=1$ [For vertical major axis]

Calculation:

  $\begin{array}{l}\frac{{\left(x+5\right)}^2}{1}+\frac{{\left(y-1\right)}^2}{9}=1\\ \frac{{\left(x+5\right)}^2}{1^2}+\frac{{\left(y-1\right)}^2}{3^2}=1\end{array}$

So, the major axis is vertical where, $h=-5,k=1,a=3,b=1$ and $c$ is obtained as,

  $\begin{array}{l}c=\sqrt{a^2-b^2}\\ c=\sqrt{3^2-1^2}\\ c=\sqrt{9-1}\\ c=\sqrt{8}=2\sqrt{2}\end{array}$

Now, the center, vertices, foci and eccentricity of an ellipse can be obtained as,

Center: $\left(h,k\right)=\left(-5,1\right)$

Vertices: $\left(h\pm a,k\right)=\left(-5,-2\right)\text{ and }\left(-5,4\right)$

Foci: $\left(h\pm c,k\right)=\left(-5,1+2\sqrt{2}\right)\text{ and }\left(-5,1-2\sqrt{2}\right)$

Eccentricity: $\frac{c}{a}=\frac{2\sqrt{2}}{3}$

The graph can be represented as,

  

Therefore, the center, vertices, foci and eccentricity of an ellipse is obtained and the graph is represented.