Chapter 8.2, Problem 46E

Solution

To calculate: The foci, vertices and eccentricity of given equation $3x^2+5y^2-12x+30y+42=0$ .

The vertices are $\left(2+\sqrt{5},-3\right),\left(2-\sqrt{5},-3\right)$ .

The foci is $\left(2-\sqrt{5},-3\right),\left(2-\sqrt{5},-3\right)$ .

The eccentricity is $\frac{\sqrt{10}}{5}$ .

Given:

A given equation is:

  $3x^2+5y^2-12x+30y+42=0$ .

Calculation:

As per the given problem,

For making the completing the square term, we have added and subtracted the term $12$ and for completing the square of $y$ , we added and subtracted the term $45$ ,

  $\begin{array}{c}3x^2+5y^2-12x+30y=-42\\ 3\left(x^2-4x+4-4\right)+5\left(y^2+6y+9-9\right)=-42\\ $ 3{(x-2)}^2-12+5{(y+3)}^2-45=-42$$ 3{(x-2)}^2+5{(y+3)}^2=-42+45+12$$ 3{(x-2)}^2+5{(y+3)}^2=15$\frac{3{(x-2)}^2}{15}+\frac{5{(y+3)}^2}{15}=\frac{15}{15}\\ \frac{{(x-2)}^2}{5}+\frac{{(y+3)}^2}{3}=1\end{array}$

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, $\left(h,k\right)=\left(2,-3\right)$ .

The values of $\left(a,b\right)=\left(\sqrt{5},\sqrt{3}\right)$

Then, the vertices will be calculated as followed:

  $\text{Vertices=}\left(\text{2+}\sqrt{\text{5}}\text{,-3}\right)\text{,}\left(\text{2-}\sqrt{\text{5}}\text{,-3}\right)$

The foci can be calculated as:

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

The value of

  $c=\sqrt{a^2-b^2}$ .

  $\begin{array}{l}c=\sqrt{5-3}\\ c=\sqrt{2}\end{array}$

Thus, the value of foci is:

  $\text{Foci=}\left(\text{2+}\sqrt{\text{2}}\text{,-3}\right)\text{,}\left(\text{2-}\sqrt{\text{2}}\text{,-3}\right)$ .

Thus, the value of eccentricity is:

  $\begin{array}{c}e=\frac{c}{b}\\ e=\frac{\sqrt{2}}{\sqrt{5}}\\ e=\frac{\sqrt{10}}{5}\end{array}$

Thus, the values are:

  $\text{Foci=}\left(\text{2+}\sqrt{\text{2}}\text{,-3}\right)\text{,}\left(\text{2-}\sqrt{\text{2}}\text{,-3}\right)$ .

The eccentricity is $=\frac{\sqrt{10}}{5}$ .

The vertices are $\left(2+\sqrt{5},-3\right),\left(2-\sqrt{5},-3\right)$ .

The graphical representation is: