Chapter EP, Problem 10.6.8EP

To determine

To write: The equation in standard form, state the graph of the equation is a parabola, circle, ellipse, or hyperbola and then graph the equation.

Answer to Problem 10.6.8EP

The standard form of equation is $\frac{{\left(x-1\right)}^2}{4^2}+\frac{{\left(y+5\right)}^2}{3^2}=1$ , the equation is Ellipse and graph is shown in figure (1).

Explanation of Solution

Given:

The equation is $9x^2-18x+16y^2+160y=-265$ .

Calculation:

Rearrange the equation in the form of perfect square as follows.

$\begin{array}{c}9x^2-18x+16y^2+160y=-265\\ 9\left(x^2-2x\right)+16\left(y^2+10y\right)=-265\\ 9\left(x^2-2x+1-1\right)+16\left(y^2+10y+25-25\right)=-265\\ 9{\left(x-1\right)}^2-9+16{\left(y+5\right)}^2-400=-265\end{array}$

Simplify further.

$\begin{array}{c}9{\left(x-1\right)}^2+16{\left(y+5\right)}^2=400+9-265\\ 9{\left(x-1\right)}^2+16{\left(y+5\right)}^2=144\end{array}$

Divide the above equation by $144$ .

$\begin{array}{c}\frac{9{\left(x-1\right)}^2}{144}+\frac{16{\left(y+5\right)}^2}{144}=\frac{144}{144}\\ \frac{{\left(x-1\right)}^2}{16}+\frac{{\left(y+5\right)}^2}{9}=1\\ \frac{{\left(x-1\right)}^2}{4^2}+\frac{{\left(y+5\right)}^2}{3^2}=1\end{array}$

The general form of Ellipse equation is $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$ .

Therefore, the given equation is Ellipse.

The graph of the equation $9x^2-18x+16y^2+160y=-265$ or $\frac{{\left(x-1\right)}^2}{4^2}+\frac{{\left(y+5\right)}^2}{3^2}=1$ is as follows.

Figure (1)

Therefore, the standard form of equation is $\frac{{\left(x-1\right)}^2}{4^2}+\frac{{\left(y+5\right)}^2}{3^2}=1$ , the equation is Ellipse and graph is shown in figure (1).