Chapter 9, Problem 29CR

To determine

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

Answer to Problem 29CR

The center of the ellipse is $\left(-2,7\right)$ , the vertices is $\left(-2\pm \frac{\sqrt{3}}{3},7\right)$ , the eccentricity is $\frac{\sqrt{10}}{4}$ and the foci is $\left(-2\pm \frac{\sqrt{30}}{12},7\right)$ . The graph is represented.

Explanation of Solution

Given information:

The expression of the ellipse is $3x^2+8y^2+12x-112y+403=0$

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:

By completing the square, write the original equation in standard form

  $\begin{array}{l}3x^2+8y^2+12x-112y+403=0\left[\text{Original equation}\right]\\ \left(3x^2+2x\right)+8\left(y^2-14y\right)=-403\left[\begin{array}{l}\text{Group terms and factor 8}\\ \text{out of }y\text{ terms}\end{array}\right]\\ 3\left(x^2+4x+4\right)+8\left(y^2-14y+49\right)=-403+12+392\left[\text{Complete the squares}\right]\\ 3{\left(x+2\right)}^2+8{\left(y-7\right)}^2=1\left[\text{Write in completed square form}\right]\\ \frac{{\left(x+2\right)}^2}{\frac{1}{3}}+\frac{{\left(y-7\right)}^2}{\frac{1}{8}}=1\left[\text{Standard form}\right]\end{array}$

So, the major axis is horizontal where, $h=-2,k=7,a=\frac{1}{3}\text{ or }\frac{\sqrt{3}}{3},b=\frac{1}{8}\text{ or }\frac{\sqrt{2}}{4}$ and $c$ is obtained as,

  $\begin{array}{l}{c}^2=a^2-b^2\\ c^2=\frac{1}{3}-\frac{1}{8}\\ c^2=\frac{5}{24}\\ c=\frac{\sqrt{30}}{12}\end{array}$

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

Center: $\left(h,k\right)=\left(-2,7\right)$

Vertices: $\left(h\pm a,k\right)=\left(-2\pm \frac{\sqrt{3}}{3},7\right)$

Foci: $\left(h\pm c,k\right)=\left(-2\pm \frac{\sqrt{30}}{12},7\right)$

Eccentricity: $\frac{c}{a}=\frac{\frac{\sqrt{30}}{12}}{\frac{\sqrt{3}}{3}}=\frac{\sqrt{10}}{4}$

The graph can be represented as,

  

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