Chapter 9, Problem 30CR

To determine

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

Answer to Problem 30CR

The center of the ellipse is $\left(1,-4\right)$ , the vertices is $\left(1,0\right)\text{,}\left(1,-8\right)$ , the eccentricity is $\frac{\sqrt{7}}{4}$ and the foci is $\left(1,-4\pm \sqrt{7}\right)$ . The graph is represented.

Explanation of Solution

Given information:

The expression of the ellipse is $16x^2+9y^2-32x+72y+16=0$

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

  $\begin{array}{l}16x^2+9y^2-32x+72y+16=0\left[\text{Original equation}\right]\\ 16\left(x^2-2x\right)+9\left(y^2+8y\right)=-16\left[\begin{array}{l}\text{Group terms and factor 16 and 9 }\\ \text{out of }x\text{ and }y\text{ terms}\end{array}\right]\\ 16\left(x^2-2x+1\right)+9\left(y^2+8y+16\right)=-16+16+144\left[\text{Complete the squares}\right]\\ 16{\left(x-1\right)}^2+9{\left(y+4\right)}^2=144\left[\text{Write in completed square form}\right]\\ \frac{{\left(x-1\right)}^2}{9}+\frac{{\left(y+4\right)}^2}{16}=1\left[\text{Standard form}\right]\end{array}$

So, the major axis is vertical where, $h=1,k=-4,a=4,b=3$ and $c$ is obtained as,

  $\begin{array}{l}c=\sqrt{a^2-b^2}\\ c=\sqrt{16-9}\\ c=\sqrt{7}\end{array}$

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

Center: $\left(h,k\right)=\left(1,-4\right)$

Vertices: $\left(h\pm a,k\right)=\left(1,0\right)\text{,}\left(1,-8\right)$

Foci: $\left(h\pm c,k\right)=\left(1,-4\pm \sqrt{7}\right)$

Eccentricity: $\frac{c}{a}=\frac{\sqrt{7}}{4}$

The graph can be represented as,

  

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