Chapter 9, Problem 34RE

a.

To determine

To find: The standard form of the equation of the ellipse

Answer to Problem 34RE

  $\frac{{\left(x-3\right)}^2}{{\left(\frac{5\sqrt{3}}{2}\right)}^2}+\frac{{\left(y+\frac{5}{2}\right)}^2}{{\left(\frac{15\sqrt{3}}{2}\right)}^2}=1$

Explanation of Solution

Given:

  $15x^2+y^2-90x+5y-140=0$

Calculation:

  $\begin{array}{c}15x^2+y^2-90x+5y-140=0\\ 15x^2-90x+y^2+5y=140\\ 15\left(x^2-6x\right)+\left(y^2+5y\right)=140\\ 15\left(x^2-6x+9-9\right)+\left(y^2+5y+\frac{25}{4}-\frac{25}{4}\right)=140\\ 15\left(x^2-6x+9\right)-90+\left(y^2+5y+\frac{25}{4}\right)-\frac{25}{4}=140\\ 15{\left(x-3\right)}^2+{\left(y+\frac{5}{2}\right)}^2=140+90+\frac{25}{4}\\ 15{\left(x-3\right)}^2+{\left(y+\frac{5}{2}\right)}^2=\frac{1125}{4}\\ \frac{15{\left(x-3\right)}^2+{\left(y+\frac{5}{2}\right)}^2}{\frac{1125}{4}}=1\\ \frac{{\left(x-3\right)}^2}{\left(\frac{75}{4}\right)}+\frac{{\left(y+\frac{5}{2}\right)}^2}{\left(\frac{1125}{4}\right)}=1\\ \frac{{\left(x-3\right)}^2}{{\left(\frac{5\sqrt{3}}{2}\right)}^2}+\frac{{\left(y+\frac{5}{2}\right)}^2}{{\left(\frac{15\sqrt{5}}{2}\right)}^2}=1\end{array}$

b.

To determine

To find: The centre, vertices, foci, and eccentricity of the ellipse $15x^2+y^2-90x+5y-140=0$ , and sketch it graph.

Answer to Problem 34RE

Center: $\left(3,-\frac{5}{2}\right)$ .

Vertices: $\left(3,-\frac{5}{2}-\frac{15\sqrt{5}}{2}\right)$ and $\left(3,-\frac{5}{2}+\frac{15\sqrt{5}}{2}\right)$ ,

Foci: $\left(3,-\frac{5}{2}-\frac{5\sqrt{42}}{2}\right)$ and $\left(3,-\frac{5}{2}+\frac{5\sqrt{42}}{2}\right)$

Eccentricity: $\text{0}\text{.966091}$

Explanation of Solution

Given:

  $15x^2+y^2-90x+5y-140=0$

Calculation:

The standard form of the given ellipse is $\frac{{\left(x-3\right)}^2}{{\left(\frac{5\sqrt{3}}{2}\right)}^2}+\frac{{\left(y+\frac{5}{2}\right)}^2}{{\left(\frac{15\sqrt{3}}{2}\right)}^2}=1$ .

Compare the given ellipse $\frac{{\left(x-3\right)}^2}{{\left(\frac{5\sqrt{3}}{2}\right)}^2}+\frac{{\left(y+\frac{5}{2}\right)}^2}{{\left(\frac{15\sqrt{3}}{2}\right)}^2}=1$ with the standard form $\frac{{\left(x-h\right)}^2}{b^2}+\frac{{\left(y-k\right)}^2}{a^2}=1$ .

This gives, $h=3,k=-\frac{5}{2},a=\frac{15\sqrt{3}}{2},b=\frac{5\sqrt{3}}{2}$ .

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

So, the vertices of the ellipse are $\left(3,-\frac{5}{2}-\frac{15\sqrt{5}}{2}\right)$ and $\left(3,-\frac{5}{2}+\frac{15\sqrt{5}}{2}\right)$ .

Now, find the value of $c$ using the equation $b^2-c^2=a^2$ .

  $\begin{array}{c}{a}^2-c^2=b^2\\ {\left(\frac{15\sqrt{3}}{2}\right)}^2-c^2={\left(\frac{5\sqrt{3}}{2}\right)}^2\\ c=\sqrt{\frac{1125}{4}-\frac{75}{4}}\\ c=\sqrt{\frac{1050}{4}}\\ c=\frac{5\sqrt{42}}{2}\end{array}$

So, the foci of the ellipse are $\left(3,-\frac{5}{2}-\frac{5\sqrt{42}}{2}\right)$ and $\left(3,-\frac{5}{2}+\frac{5\sqrt{42}}{2}\right)$ .

Then, the value of the eccentricity is $e=\frac{c}{a}=\frac{\left(\frac{5\sqrt{42}}{2}\right)}{\left(\frac{15\sqrt{5}}{2}\right)}=\frac{\sqrt{42}}{3\sqrt{5}}=\text{0}\text{.966091}$ .

c.

To determine

To sketch: The graph of the ellipse $15x^2+y^2-90x+5y-140=0$ .

Explanation of Solution

Given:

  $15x^2+y^2-90x+5y-140=0$

Graph:

The graph of the ellipse can be drawn as shown below.