Chapter 9.2, Problem 43E

(a)

To determine

To find: The standard form of the equation of ellipse $12x^2+20y^2-12x+40y-37=0$ .

Answer to Problem 43E

The standard form of the equation of ellipse $12x^2+20y^2-12x+40y-37=0$ is $\frac{{\left(x-\frac{1}{2}\right)}^2}{{\left(\sqrt{5}\right)}^2}+\frac{{\left(y+1\right)}^2}{{\left(\sqrt{3}\right)}^2}=1$ .

Explanation of Solution

Given information:

The equation of the ellipse is $12x^2+20y^2-12x+40y-37=0$ .

Calculation:

Rewrite the given equation of ellipse.

  $\begin{array}{c}12x^2+20y^2-12x+40y-37=0\\ 12\left(x^2-x\right)+20\left(y^2+2y\right)-37=0\\ 12\left(x^2-x+\frac{1}{4}-\frac{1}{4}\right)+20\left(y^2+2y+1-1\right)-37=0\\ 12{\left(x-\frac{1}{2}\right)}^2+20{\left(y+1\right)}^2-3-20-37=0\end{array}$

Further simplify,

  $\begin{array}{c}12{\left(x-\frac{1}{2}\right)}^2+20{\left(y+1\right)}^2=60\\ \frac{12{\left(x-\frac{1}{2}\right)}^2}{60}+\frac{20{\left(y+1\right)}^2}{60}=1\\ \frac{{\left(x-\frac{1}{2}\right)}^2}{5}+\frac{{\left(y+1\right)}^2}{3}=1\\ \frac{{\left(x-\frac{1}{2}\right)}^2}{{\left(\sqrt{5}\right)}^2}+\frac{{\left(y+1\right)}^2}{{\left(\sqrt{3}\right)}^2}=1\end{array}$

Therefore, the standard form of the equation of ellipse $12x^2+20y^2-12x+40y-37=0$ is $\frac{{\left(x-\frac{1}{2}\right)}^2}{{\left(\sqrt{5}\right)}^2}+\frac{{\left(y+1\right)}^2}{{\left(\sqrt{3}\right)}^2}=1$ .

(b)

To determine

To find: The center, vertices, foci and eccentricity of the ellipse $12x^2+20y^2-12x+40y-37=0$ .

Answer to Problem 43E

The center, vertices, foci and eccentricity of the ellipse $12x^2+20y^2-12x+40y-37=0$ are $\left(\frac{1}{2},-1\right)$ , $\left(2.74,-1\right),\left(-1.74,-1\right)$ , $\left(1.91,-1\right),\left(-0.91,-1\right)$ and $0.63$ respectively.

Explanation of Solution

Given information:

The equation of the ellipse is $12x^2+20y^2-12x+40y-37=0$ .

Calculation:

As calculated in part(a), the standard form of the equation of ellipse is $\frac{{\left(x-\frac{1}{2}\right)}^2}{{\left(\sqrt{5}\right)}^2}+\frac{{\left(y+1\right)}^2}{{\left(\sqrt{3}\right)}^2}=1$ .

Compare the equation of ellipse $\frac{{\left(x-\frac{1}{2}\right)}^2}{{\left(\sqrt{5}\right)}^2}+\frac{{\left(y+1\right)}^2}{{\left(\sqrt{3}\right)}^2}=1$ with general equation $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$ .

This gives $a=\sqrt{5}$ , $b=\sqrt{3}$ , $h=\frac{1}{2}$ and $k=-1$ . The major axis of the ellipse is horizontal.

Use the formula.

  $\begin{array}{c}{c}^2=a^2-b^2\\ c^2={\left(\sqrt{5}\right)}^2-{\left(\sqrt{3}\right)}^2\\ c^2=5-3\\ c=\sqrt{2}\\ \approx 1.41\end{array}$

Use the formula for eccentricity.

  $\begin{array}{c}e=\frac{c}{a}\\ =\frac{\sqrt{2}}{\sqrt{5}}\\ \approx 0.63\end{array}$

The center of the ellipse is $\left(h,k\right)=\left(\frac{1}{2},-1\right)$ .

The vertices of the ellipse are:

  $\begin{array}{c}\left(h\pm a,k\right)=\left(\frac{1}{2}\pm \sqrt{5},-1\right)\\ =\left(2.74,-1\right),\left(-1.74,-1\right)\end{array}$

The foci of the ellipse are:

  $\begin{array}{c}\left(h\pm c,k\right)=\left(\frac{1}{2}\pm 1.41,-1\right)\\ =\left(1.91,-1\right),\left(-0.91,-1\right)\end{array}$

The eccentricity of ellipse is $e=0.63$ .

Therefore, The center, vertices, foci and eccentricity of the ellipse $12x^2+20y^2-12x+40y-37=0$ are $\left(\frac{1}{2},-1\right)$ , $\left(2.74,-1\right),\left(-1.74,-1\right)$ , $\left(1.91,-1\right),\left(-0.91,-1\right)$ and $0.63$ respectively.

(c)

To determine

To graph: The ellipse of equation $12x^2+20y^2-12x+40y-37=0$ .

Answer to Problem 43E

The graph of the ellipse $12x^2+20y^2-12x+40y-37=0$ is shown in Figure(1).

Explanation of Solution

Given information:

The equation of the ellipse is $12x^2+20y^2-12x+40y-37=0$ .

Calculation:

The graph of the ellipse $12x^2+20y^2-12x+40y-37=0$ is shown below.

  

Figure (1)

The graph of the equation $12x^2+20y^2-12x+40y-37=0$ with the help of graphing utility:

  

Figure (2)

As both the graphs gives the same values at each point.

Therefore, both the graphs are similar.