Chapter 9.1, Problem 67E

To determine

To find the Vertex, Focus and Directrix of a parabola.

Answer to Problem 67E

The values of $\text{vertex}=\left(-2,1\right),\text{focus}=\left(-2,-\frac{1}{2}\right)\text{ and directrix }\left(x=\frac{5}{2}\right)$

Explanation of Solution

Given information:

The equation is

  $x^2+4x+6y-2=0$

Calculation:

The Vertex, Focus and Directrix of a parabola can be obtained as

First convert the equation to standard form by completing the square.

  $\begin{array}{l}{x}^2+4x+6y-2=0\\ x^2+4x=-6y+2\\ x^2+4x+4=-6y+2+4\left[\text{add 4 both sides}\right]\\ {\left(x+2\right)}^2=-6y+6\\ {\left(x+2\right)}^2\text{=}-6\left(y-1\right)\left[\text{standard form}\right]\end{array}$

The standard form of the equation of a parabola with vertex at $\left(h,k\right)$ is

  ${\left(x-h\right)}^2=4p\left(y-k\right)$

Comparing the given equation to the standard form of the equation it can observed that the vertex is

  $h=-2\text{, }k=1\text{ and }p=-\frac{3}{2}$

Add $p\text{ to }k$ to find the focus

  $\begin{array}{l}\text{focus = }\left(h,k+p\right)\\ \left(h,k+p\right)=\left(-2,1+\left(-\frac{3}{2}\right)\right)\left[\text{since }h=-2,k=1,p=-\frac{3}{2}\right]\\ \left(h,k+p\right)=\left(-2,1-\frac{3}{2}\right)\\ \left(h,k+p\right)=\left(-2,-\frac{1}{2}\right)\\ \left(h,k+p\right)=\left(-2,-\frac{1}{2}\right)\end{array}$

Subtract $p\text{ to }k$ to find the directrix

  $\begin{array}{l}\text{Directrix, }\left(y=k-p\right)\\ \left(y=1-\left(-\frac{3}{2}\right)\right)\\ \left(y=1+\frac{3}{2}\right)\\ y=\left(\frac{5}{2}\right)\end{array}$

Because p is negative, the parabola opens downwords and the vertex of the parabola will be at $\left(-2,1\right)$ and the focus is at $\left(-2,-\frac{1}{2}\right)$ .

So, the graph is obtained as:

  

Hence,

The values of $\text{vertex}=\left(-2,1\right),\text{focus}=\left(-2,-\frac{1}{2}\right)\text{ and directrix }\left(x=\frac{5}{2}\right)$

To verify the above graph, use graphing calculator to graph the given equation follow the steps given below:

Step 1: Press WINDOW button to access the Window editor.

Step 2: Press $\boxed{\text{Y=}}$ button.

Step 3: Enter the expression ${\left(x+2\right)}^2\text{=}-6\left(y-1\right)$ which is required to graph.

Step 4: Press GRAPH button to graph the function and adjust the windows according to the graph.

The graph is obtained as:

  

Hence, the graph is verified.