Chapter 9.1, Problem 73E

To determine

To find: The vertex, focus and directrix of the parabola $x^2+4x+6y-2=0$ and verify with the help of graphing utility.

Answer to Problem 73E

The vertex, focus and directrix of the parabola $x^2+4x+6y-2=0$ are $\left(-2,1\right)$ , $\left(-2,-\frac{1}{2}\right)$ , $y=\frac{5}{2}$ respectively. The graph of the parabola with vertex, focus and directrix is shown in Figure(1).

Explanation of Solution

Given information:

The equation of the parabola is $x^2+4x+6y-2=0$ .

Calculation:

Simplify the equation of parabola.

  $\begin{array}{c}{x}^2+4x+6y-2=0\\ x^2+4x+2^2-2^2+6y-2=0\\ {\left(x+2\right)}^2+6y-4-2=0\\ {\left(x+2\right)}^2=-6y+6\\ {\left(x+2\right)}^2=-6\left(y-1\right)\end{array}$

Compare the given equation of parabola with general vertical axis parabola equation ${\left(x-h\right)}^2=4p\left(y-k\right)$ . This gives the value of $h$ , $k$ , $p$ as $-2$ , $1$ , $-\frac{3}{2}$ respectively.

The vertex of the parabola $x^2+4x+6y-2=0$ is $\left(h,k\right)=\left(-2,1\right)$ .

The focus of the horizontal axis parabola $x^2+4x+6y-2=0$ is:

  $\begin{array}{c}\left(h,k+p\right)=\left(-2,1+\left(-\frac{3}{2}\right)\right)\\ =\left(-2,-\frac{1}{2}\right)\end{array}$

The directrix of the parabola $x^2+4x+6y-2=0$ is:

  $\begin{array}{c}y=k-p\\ =1-\left(-\frac{3}{2}\right)\\ =\frac{5}{2}\end{array}$

Therefore, The vertex, focus and directrix of the parabola $x^2+4x+6y-2=0$ are $\left(-2,1\right)$ , $\left(-2,-\frac{1}{2}\right)$ , $y=\frac{5}{2}$ respectively.

The graph of the parabola $x^2+4x+6y-2=0$ with vertex, focus and directrix is shown below.

  

Figure (1)

Here we can verify the vertex, focus, directrix of the parabola $x^2+4x+6y-2=0$ are $\left(-2,1\right)$ , $\left(-2,-\frac{1}{2}\right)$ , $y=\frac{5}{2}$ respectively.