Chapter 9.2, Problem 41E

(a)

To determine

To find: The standard form ellipse for the given equation $6x^2+2y^2+18x-10y+2=0$ .

Answer to Problem 41E

The standard form of the equation of ellipse $6x^2+2y^2+18x-10y+2=0$ is $\frac{{\left(x+\frac{3}{2}\right)}^2}{2^2}+\frac{{\left(y-\frac{5}{2}\right)}^2}{{\left(2\sqrt{3}\right)}^2}=1$ .

Explanation of Solution

Given information:

The equation of the ellipse is $6x^2+2y^2+18x-10y+2=0$ .

Calculation:

Rewrite the given equation of ellipse.

  $\begin{array}{c}6x^2+2y^2+18x-10y+2=0\\ 6x^2+18x+2y^2-10y+2=0\\ 6\left(x^2+3x\right)+2\left(y^2-5y\right)+2=0\\ 6\left(x^2+3x+\frac{9}{4}-\frac{9}{4}\right)+2\left(y^2-5y+\frac{25}{4}-\frac{25}{4}\right)+2=0\end{array}$

Further simplify,

  $\begin{array}{c}6{\left(x+\frac{3}{2}\right)}^2+2{\left(y-\frac{5}{2}\right)}^2-\frac{27}{2}-\frac{25}{2}+2=0\\ 6{\left(x+\frac{3}{2}\right)}^2+2{\left(y-\frac{5}{2}\right)}^2=24\\ \frac{6{\left(x+\frac{3}{2}\right)}^2}{24}+\frac{2{\left(y-\frac{5}{2}\right)}^2}{24}=1\end{array}$

Simplify further,

  $\begin{array}{c}\frac{{\left(x+\frac{3}{2}\right)}^2}{4}+\frac{{\left(y-\frac{5}{2}\right)}^2}{12}=1\\ \frac{{\left(x+\frac{3}{2}\right)}^2}{2^2}+\frac{{\left(y-\frac{5}{2}\right)}^2}{{\left(2\sqrt{3}\right)}^2}=1\end{array}$

Therefore, the standard form of the equation of ellipse $6x^2+2y^2+18x-10y+2=0$ is $\frac{{\left(x+\frac{3}{2}\right)}^2}{2^2}+\frac{{\left(y-\frac{5}{2}\right)}^2}{{\left(2\sqrt{3}\right)}^2}=1$ .

(b)

To determine

To find: The center, vertices, foci and eccentricity of the ellipse $6x^2+2y^2+18x-10y+2=0$ .

Answer to Problem 41E

The center, vertices, foci and eccentricity of the ellipse $6x^2+2y^2+18x-10y+2=0$ are $\left(-\frac{3}{2},\frac{5}{2}\right)$ , $\left(-1.5,-0.96\right),\left(-1.5,5.964\right)$ , $\left(-1.5,-0.33\right),\left(-1.5,5.33\right)$ and $0.82$ respectively.

Explanation of Solution

Given information:

The equation of the ellipse is $6x^2+2y^2+18x-10y+2=0$ .

Calculation:

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

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

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

Use the formula.

  $\begin{array}{c}{c}^2=a^2-b^2\\ c^2={\left(2\sqrt{3}\right)}^2-2^2\\ c^2=12-4\\ c\approx 2.83\end{array}$

Use the formula for eccentricity.

  $\begin{array}{c}e=\frac{c}{a}\\ =\frac{2\sqrt{2}}{2\sqrt{3}}\\ \approx 0.82\end{array}$

The center of the ellipse is $\left(h,k\right)=\left(-\frac{3}{2},\frac{5}{2}\right)$ .

The vertices of the ellipse are:

  $\begin{array}{c}\left(h,k\pm a\right)=\left(-\frac{3}{2},\frac{5}{2}\pm 2\sqrt{3}\right)\\ =\left(-1.5,-0.96\right),\left(-1.5,5.964\right)\end{array}$

The foci of the ellipse are:

  $\begin{array}{c}\left(h,k\pm c\right)=\left(-\frac{3}{2},\frac{5}{2}\pm 2\sqrt{2}\right)\\ =\left(-1.5,-0.33\right),\left(-1.5,5.33\right)\end{array}$

The eccentricity of ellipse is $e=0.82$ .

Therefore, The center, vertices, foci and eccentricity of the ellipse $6x^2+2y^2+18x-10y+2=0$ are $\left(-\frac{3}{2},\frac{5}{2}\right)$ , $\left(-1.5,-0.96\right),\left(-1.5,5.964\right)$ , $\left(-1.5,-0.33\right),\left(-1.5,5.33\right)$ and $0.82$ respectively.

(c)

To determine

To graph: The ellipse of equation $6x^2+2y^2+18x-10y+2=0$ .

Answer to Problem 41E

The graph of the ellipse $6x^2+2y^2+18x-10y+2=0$ is shown in Figure (1).

Explanation of Solution

Given information:

The equation of the ellipse is $6x^2+2y^2+18x-10y+2=0$ .

Calculation:

The graph of the ellipse $6x^2+2y^2+18x-10y+2=0$ is shown below.

  

Figure (1)

The graph of the equation $6x^2+2y^2+18x-10y+2=0$ 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.