Chapter EP, Problem 10.6.5EP

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 10.6.5EP

The standard form of equation is $x=2\left(y-{\left(-3\right)}^2\right)+\left(-12\right)$ , the equation is Parabola and graph is shown in figure (1).

Explanation of Solution

Given:

The equation is $2y^2+12y-x+6=0$ .

Calculation:

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

$\begin{array}{c}2y^2+12y-x+6=0\\ 2\left(y^2+6y\right)=x-6\\ 2\left(y^2+6y+9-9\right)=x-6\\ 2{\left(y+3\right)}^2-18=\left(x-6\right)\end{array}$

Simplify further.

$\begin{array}{c}x=2{\left(y+3\right)}^2-18+6\\ x=2\left(y-{\left(-3\right)}^2\right)+\left(-12\right)\end{array}$

The general form of Parabola equation is $x=a\left(y-k^2\right)+h$ .

Therefore, the given equation is Parabola.

The graph of the equation $2y^2+12y-x+6=0$ or $x=2\left(y-{\left(-3\right)}^2\right)+\left(-12\right)$ is as follows.

Figure (1)

Therefore, the standard form of equation is $x=2\left(y-{\left(-3\right)}^2\right)+\left(-12\right)$ , the equation is Parabola and graph is shown in figure (1).