Chapter 9.2, Problem 40E

(a)

To determine

To find:The standard form of the equation of ellipse $9x^2+4y^2-54x+40y+37=0$ .

Answer to Problem 40E

The standard form of the equation of ellipse $9x^2+4y^2-54x+40y+37=0$ is $\frac{{\left(x-3\right)}^2}{4^2}+\frac{{\left(y+5\right)}^2}{6^2}=1$ .

Explanation of Solution

Given information:

The equation of the ellipse is $9x^2+4y^2-54x+40y+37=0$ .

Calculation:

Rewrite the given equation of ellipse.

  $\begin{array}{c}9x^2+4y^2-54x+40y+37=0\\ 9x^2-54x+4y^2+40y+37=0\\ 9\left(x^2-6x\right)+4\left(y^2+10y\right)+37=0\\ 9\left(x^2-6x+9-9\right)+4\left(y^2+10y+25-25\right)+37=0\\ 9\left(x^2-6x+9\right)+4\left(y^2+10y+25\right)+37-100-81=0\end{array}$

Further simplify,

  $\begin{array}{c}9\left(x^2-6x+9\right)+4\left(y^2+10y+25\right)+37-100-81=0\\ 9{\left(x-3\right)}^2+4{\left(y+5\right)}^2=144\\ \frac{9{\left(x-3\right)}^2}{144}+\frac{4{\left(y+5\right)}^2}{144}=1\\ \frac{{\left(x-3\right)}^2}{16}+\frac{{\left(y+5\right)}^2}{36}=1\\ \frac{{\left(x-3\right)}^2}{4^2}+\frac{{\left(y+5\right)}^2}{6^2}=1\end{array}$

Therefore, the standard form of the equation of ellipse $9x^2+4y^2-54x+40y+37=0$ is $\frac{{\left(x-3\right)}^2}{4^2}+\frac{{\left(y+5\right)}^2}{6^2}=1$ .

(b)

To determine

To find:The center, vertices, foci and eccentricity of the ellipse $9x^2+4y^2-54x+40y+37=0$ .

Answer to Problem 40E

The center, vertices, foci and eccentricity of the ellipse $9x^2+4y^2-54x+40y+37=0$ are $\left(3,-5\right)$ , $\left(3,1\right),\left(3,-11\right)$ , $\left(3,-0.53\right),\left(3,-9.47\right)$ and $0.74$ respectively.

Explanation of Solution

Given information:

The equation of the ellipse is $9x^2+4y^2-54x+40y+37=0$ .

Calculation:

As calculated in part(a), the standard form of the equation of ellipse is $\frac{{\left(x-3\right)}^2}{4^2}+\frac{{\left(y+5\right)}^2}{6^2}=1$ .

Compare the equation of ellipse $\frac{{\left(x-3\right)}^2}{4^2}+\frac{{\left(y+5\right)}^2}{6^2}=1$ with general equation $\frac{{\left(x-h\right)}^2}{b^2}+\frac{{\left(y-k\right)}^2}{a^2}=1$ .

This gives $a=6$ , $b=4$ , $h=3$ and $k=-5$ . The major axis of the ellipse is vertical.

Use the formula.

  $\begin{array}{c}{c}^2=a^2-b^2\\ c^2=36-16\\ c^2=20\\ c=2\sqrt{5}\\ \approx 4.47\end{array}$

Use the formula for eccentricity.

  $\begin{array}{c}e=\frac{c}{a}\\ =\frac{2\sqrt{5}}{6}\\ \approx 0.74\end{array}$

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

The vertices of the ellipse are:

  $\begin{array}{c}\left(h,k\pm a\right)=\left(3,-5\pm 6\right)\\ =\left(3,1\right),\left(3,-11\right)\end{array}$

The foci of the ellipse are:

  $\begin{array}{c}\left(h,k\pm c\right)=\left(3,-5\pm 2\sqrt{5}\right)\\ =\left(3,-0.53\right),\left(3,-9.47\right)\end{array}$

The eccentricity of ellipse is $e=0.74$ .

Therefore, The center, vertices, foci and eccentricity of the ellipse $9x^2+4y^2-54x+40y+37=0$ are $\left(3,-5\right)$ , $\left(3,1\right),\left(3,-11\right)$ , $\left(3,-0.53\right),\left(3,-9.47\right)$ and $0.74$ respectively.

(c)

To determine

To graph:The ellipse of equation $9x^2+4y^2-54x+40y+37=0$ .

Answer to Problem 40E

The graph of the ellipse $9x^2+4y^2-54x+40y+37=0$ is shown in Figure(1).

Explanation of Solution

Given information:

The equation of the ellipse is $9x^2+4y^2-54x+40y+37=0$ .

Calculation:

The graph of the ellipse $9x^2+4y^2-54x+40y+37=0$ is shown below.

  

Figure(1)

Verify the result graphically:

The graph of the equation $9x^2+4y^2-54x+40y+37=0$ 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.