Chapter 10.6, Problem 19E

To determine

To identify: the conic section represented by each equation; write the equation in standard form and then graph the equation.

Answer to Problem 19E

Ellipse

  $\frac{{\left(x-4\right)}^2}{2^2}+\frac{{\left(y-5\right)}^2}{{\left(\sqrt{10}\right)}^2}=1$

Explanation of Solution

Given information:

  $5x^2+2y^2-40x-20y+110=0$

Calculation:

Rewrite the equation in the general form $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ to derive the coefficients as shown below.

  $\begin{array}{l}5x^2+0xy+2y^2+\left(-40\right)x+\left(-20\right)y+110=0\\ \therefore A=5\text{ and }C=2\end{array}$

Since $A\text{ and }C$ have same sign but different values. Thus, the conic section represented by the equation is an ellipse.

The standard equation of an ellipse is shown below.

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

Now rewrite the equation in the standard form as follows:

  $\begin{array}{c}5x^2+2y^2-40x-20y+110=0\\ 5x^2-40x+2y^2-20y=-110\left[\text{Rewrite the equation}\right]\\ 5\left(x^2-8x\right)+2\left(y^2-10y\right)=-110\left[\text{Taking GCF}\right]\\ 5\left(x^2-8x+4\right)+2\left(y^2-10y+5\right)=-110\left[\text{Completing the square}\right]\\ {\left(\sqrt{10}\right)}^2{\left(x-4\right)}^2+2^2{\left(y-5\right)}^2=40\left[\text{Factors}\right]\\ \frac{{\left(x-4\right)}^2}{2^2}+\frac{{\left(y-5\right)}^2}{{\left(\sqrt{10}\right)}^2}=1\left[\text{Divide each side by 40}\right]\\ \frac{{\left(x-4\right)}^2}{2^2}+\frac{{\left(y-5\right)}^2}{{\left(\sqrt{10}\right)}^2}=1\left[\text{Standard equation}\right]\end{array}$

Thus, the standard equation of ellipse is $\frac{{\left(x-4\right)}^2}{2^2}+\frac{{\left(y-5\right)}^2}{{\left(\sqrt{10}\right)}^2}=1$ .

The value of $a,b,c,h\text{ and }k$ are shown below.

  $\begin{array}{l}a=2\\ b=\sqrt{10}\\ c=\sqrt{b^2-a^2}=\sqrt{6}\\ h=4\\ k=5\end{array}$

The center $\left(h,k\right)$ is $\left(4,5\right)$ , the foci $\left(h\pm c,k\right)$ is $\left(4\pm \sqrt{6},5\right)$ , major axis vertices $\left(h\pm a,k\right)$ are $\left(6,5\right)\text{ and }\left(2,5\right)$ , and the minor axis vertices $\left(h,k\pm b\right)$ are $\left(4,5+\sqrt{10}\right)\text{ and }\left(4,5-\sqrt{10}\right)$ .Thus, the graph of the given equation is shown below.