Chapter 9.2, Problem 42E

(a)

To determine

To find: The standard form of the equation of ellipse $x^2+4y^2-6x+20y-2=0$ .

Answer to Problem 42E

The standard form of the equation of ellipse $x^2+4y^2-6x+20y-2=0$ is $\frac{{\left(x-3\right)}^2}{6^2}+\frac{{\left(y+\frac{5}{2}\right)}^2}{3^2}=1$ .

Explanation of Solution

Given information:

The equation of the ellipse is $x^2+4y^2-6x+20y-2=0$ .

Calculation:

Rewrite the given equation of ellipse.

  $\begin{array}{c}{x}^2+4y^2-6x+20y-2=0\\ x^2-6x+4y^2+20y-2=0\\ \left(x^2-6x\right)+4\left(y^2+5y\right)-2=0\\ \left(x^2-6x+9-9\right)+4\left(y^2+5y+\frac{25}{4}-\frac{25}{4}\right)-2=0\end{array}$

Further simplify,

  $\begin{array}{c}{\left(x-3\right)}^2+4{\left(y+\frac{5}{2}\right)}^2-9-25-2=0\\ {\left(x-3\right)}^2+4{\left(y+\frac{5}{2}\right)}^2=36\\ \frac{{\left(x-3\right)}^2}{36}+\frac{4{\left(y+\frac{5}{2}\right)}^2}{36}=1\\ \frac{{\left(x-3\right)}^2}{6^2}+\frac{{\left(y+\frac{5}{2}\right)}^2}{3^2}=1\end{array}$

Therefore, the standard form of the equation of ellipse $x^2+4y^2-6x+20y-2=0$ is $\frac{{\left(x-3\right)}^2}{6^2}+\frac{{\left(y+\frac{5}{2}\right)}^2}{3^2}=1$ .

(b)

To determine

To find: The center, vertices, foci and eccentricity of the ellipse $x^2+4y^2-6x+20y-2=0$ .

Answer to Problem 42E

The center, vertices, foci and eccentricity of the ellipse $x^2+4y^2-6x+20y-2=0$ are $\left(3,-\frac{5}{2}\right)$ , $\left(9,-2.5\right),\left(-3,-2.5\right)$ , $\left(8.2,-2.5\right),\left(-2.2,-2.5\right)$ and $0.87$ respectively.

Explanation of Solution

Given information:

The equation of the ellipse is $x^2+4y^2-6x+20y-2=0$ .

Calculation:

As calculated in part(a), the standard form of the equation of ellipse is $\frac{{\left(x-3\right)}^2}{6^2}+\frac{{\left(y+\frac{5}{2}\right)}^2}{3^2}=1$ .

Compare the equation of ellipse $\frac{{\left(x-3\right)}^2}{6^2}+\frac{{\left(y+\frac{5}{2}\right)}^2}{3^2}=1$ with general equation $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$ .

This gives $a=6$ , $b=3$ , $h=3$ and $k=-\frac{5}{2}$ . The major axis of the ellipse is horizontal.

Use the formula.

  $\begin{array}{c}{c}^2=a^2-b^2\\ c^2={\left(6\right)}^2-3^2\\ c^2=36-9\\ c=3\sqrt{3}\\ \approx 5.20\end{array}$

Use the formula for eccentricity.

  $\begin{array}{c}e=\frac{c}{a}\\ =\frac{5.20}{6}\\ \approx 0.87\end{array}$

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

The vertices of the ellipse are:

  $\begin{array}{c}\left(h\pm a,k\right)=\left(3\pm 6,-\frac{5}{2}\right)\\ =\left(9,-2.5\right),\left(-3,-2.5\right)\end{array}$

The foci of the ellipse are:

  $\begin{array}{c}\left(h\pm c,k\right)=\left(3\pm 5.20,-\frac{5}{2}\right)\\ =\left(8.2,-2.5\right),\left(-2.2,-2.5\right)\end{array}$

The eccentricity of ellipse is $e=0.87$ .

Therefore, The center, vertices, foci and eccentricity of the ellipse $x^2+4y^2-6x+20y-2=0$ are $\left(3,-\frac{5}{2}\right)$ , $\left(9,-2.5\right),\left(-3,-2.5\right)$ , $\left(8.2,-2.5\right),\left(-2.2,-2.5\right)$ and $0.87$ respectively.

(c)

To determine

To graph: The ellipse of equation $x^2+4y^2-6x+20y-2=0$ .

Answer to Problem 42E

The graph of the ellipse $x^2+4y^2-6x+20y-2=0$ is shown in Figure (1).

Explanation of Solution

Given information:

The equation of the ellipse is $x^2+4y^2-6x+20y-2=0$ .

Calculation:

The graph of the ellipse $x^2+4y^2-6x+20y-2=0$ is shown below.

  

Figure (1)

The graph of the equation $x^2+4y^2-6x+20y-2=0$ 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.