Chapter 8.2, Problem 47E

Solution

To calculate: The foci, vertices and eccentricity of given equation $9x^2+16y^2+54x-32y-47=0$ .

The vertices is $\left(-3,3\right),\left(-3,-3\right)$ .

The foci is $\left(-3+\sqrt{7},1\right),\left(-3-\sqrt{7},1\right)$ .

The eccentricity is $\frac{\sqrt{7}}{4}$ .

Given:

A given equation is:

  $9x^2+16y^2+54x-32y-47=0$ .

Calculation:

As per the given problem,

For making the completing the square term, we have added and subtracted the term $81$ and for completing the square of $y$ , we added and subtracted the term $16$ ,

  $\begin{array}{c}9x^2+16y^2+54x-32y=47\\ 9\left(x^2+6x+9-9\right)+16\left(y^2-2y+1-1\right)=47\\ $ 9{(x+3)}^2-81+16{(y-1)}^2-16=47$$ 9{(x+3)}^2+16{(y-1)}^2=47+16+81$$ 9{(x+3)}^2+16{(y-1)}^2=144$\frac{9{(x+3)}^2}{144}+\frac{16{(y-1)}^2}{144}=\frac{144}{144}\\ \frac{{(x+3)}^2}{16}+\frac{{(y-1)}^2}{9}=1\end{array}$

The general form of ellipse is:

  $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$

The vertices in given general form can be written as:

Vertices $=\left(h+a,k\right),\left(h-a,k\right)$ .

Here, $\left(h,k\right)=\left(-3,1\right)$ .

The values of $\left(a,b\right)=\left(4,3\right)$

Then, the vertices will be calculated as followed:

  $\begin{array}{c}\text{Vertices=}\left(\text{-3+4,1}\right)\text{,}\left(\text{-3-4,1}\right)\\ \text{=}\left(\text{1,1}\right)\text{,}\left(\text{-7,1}\right)\end{array}$

The foci can be calculated as:

Foci $=\left(h,k+c\right),\left(h,k-c\right)$ .

The value of

  $c=\sqrt{a^2-b^2}$ .

  $\begin{array}{l}c=\sqrt{16-9}\\ c=\sqrt{7}\end{array}$

Thus, the value of foci is:

  $\text{Foci=}\left(\text{-3,1+}\sqrt{\text{7}}\right)\text{,}\left(\text{-3,1-}\sqrt{\text{7}}\right)$ .

Thus, the value of eccentricity is:

  $\begin{array}{c}e=\frac{c}{b}\\ e=\frac{\sqrt{7}}{3}\end{array}$

Thus, the values are:

  $\text{Foci=}\left(\text{-3,1+}\sqrt{\text{7}}\right)\text{,}\left(\text{-3,1-}\sqrt{\text{7}}\right)$ .

  $\text{Vertices=}\left(\text{1,1}\right)\text{,}\left(\text{-7,1}\right)$ .

  $\text{eccentricity=}\frac{\sqrt{\text{7}}}{\text{4}}$ .

Using the graphing calculator,