Lesson Objectives

- Learn how to find the focus of a parabola
- Learn how to find the directrix of a parabola
- Learn how to write a parabola in vertex form

## Parabolas - Finding the focus, directrix, and equation

In this lesson, we want to explore the geometric definition of a parabola. A parabola is the set of all points in a plane equidistant from a fixed point, known as the focus, and a fixed line, known as the directrix. In this section, we will often be asked to find the focus, the directrix, or write the equation of the parabola. In the past, we saw the parabola written as:

Vertical: $$(x - h)^2=4p(y - k)$$ Horizontal: $$(y - k)^2=4p(x - h)$$ Our focus or fixed point is distance |p| from the vertex (h, k).

Let's look at an example.

Example #1: Write the vertex form of the parabola. $$\text{Vertex}:(-9, -2)$$ $$\text{Focus}:\left(-9, -\frac{5}{4}\right)$$ First, we want to determine if this is a vertical or horizontal parabola. Since the x-location of the vertex and the focus are the same, this means we will have a vertical parabola. Let's use that formula: $$(x - h)^2=4p(y - k)$$ Plug in for h and k: $$(x + 9)^2=4p(y + 2)$$ Recall that the focus has a distance of |p| units from the vertex.

If we go from the vertex to the focus, we can see we need to travel up by 3/4 units. This tells us that p = 3/4. Let's plug in: $$(x + 9)^2=4 \cdot \frac{3}{4}(y + 2)$$ To put this in vertex form, let's solve for y: $$(x + 9)^2=3(y + 2)$$ $$(x + 9)^2=3y + 6$$ $$3y + 6=(x + 9)^2$$ $$3y=(x + 9)^2 - 6$$ $$y=\frac{1}{3}(x + 9)^2 - 2$$ Now, if you are asked to find the directrix, recall that the distance from any point to the focus is the same as the distance from that same point to the directrix. So from the vertex, we traveled 3/4 units up to get to the focus. This means we would travel 3/4 units down to get to the directrix. The x-value would be the same, so just take the y-value from the vertex, which is -2 and subtract away 3/4. This gives us: $$-2 - \frac{3}{4}$$ $$-\frac{8}{4}- \frac{3}{4}=-\frac{11}{4}$$ The directrix: $$y=-\frac{11}{4}$$

### Standard Form

Vertical: $$y=ax^2 + bx + c$$ Horizontal: $$x=ay^2 + by + c$$### Vertex Form

Vertical: $$y=a(x - h)^2 + k$$ Horizontal: $$x=a(y - k)^2 + h$$ Now, we will see yet another form. A parabola with a vertex (h,k) can also be written as:Vertical: $$(x - h)^2=4p(y - k)$$ Horizontal: $$(y - k)^2=4p(x - h)$$ Our focus or fixed point is distance |p| from the vertex (h, k).

Let's look at an example.

Example #1: Write the vertex form of the parabola. $$\text{Vertex}:(-9, -2)$$ $$\text{Focus}:\left(-9, -\frac{5}{4}\right)$$ First, we want to determine if this is a vertical or horizontal parabola. Since the x-location of the vertex and the focus are the same, this means we will have a vertical parabola. Let's use that formula: $$(x - h)^2=4p(y - k)$$ Plug in for h and k: $$(x + 9)^2=4p(y + 2)$$ Recall that the focus has a distance of |p| units from the vertex.

If we go from the vertex to the focus, we can see we need to travel up by 3/4 units. This tells us that p = 3/4. Let's plug in: $$(x + 9)^2=4 \cdot \frac{3}{4}(y + 2)$$ To put this in vertex form, let's solve for y: $$(x + 9)^2=3(y + 2)$$ $$(x + 9)^2=3y + 6$$ $$3y + 6=(x + 9)^2$$ $$3y=(x + 9)^2 - 6$$ $$y=\frac{1}{3}(x + 9)^2 - 2$$ Now, if you are asked to find the directrix, recall that the distance from any point to the focus is the same as the distance from that same point to the directrix. So from the vertex, we traveled 3/4 units up to get to the focus. This means we would travel 3/4 units down to get to the directrix. The x-value would be the same, so just take the y-value from the vertex, which is -2 and subtract away 3/4. This gives us: $$-2 - \frac{3}{4}$$ $$-\frac{8}{4}- \frac{3}{4}=-\frac{11}{4}$$ The directrix: $$y=-\frac{11}{4}$$

#### Skills Check:

Example #1

Find the vertex form. $$\text{Vertex}:(-5, 1)$$ $$\text{Focus}:\left(-5, \frac{3}{4}\right)$$

Please choose the best answer.

A

$$y=(x + 5)^2 - 1$$

B

$$y=-(x + 5)^2 + 1$$

C

$$y=(x + 5)^2 + 1$$

D

$$y=-2(x + 1)^2 - 5$$

E

$$y=\frac{1}{2}(x - 3)^2 + 7$$

Example #2

Find the vertex form. $$\text{Vertex}:(-3, 7)$$ $$\text{Focus}:\left(-3, \frac{57}{8}\right)$$

Please choose the best answer.

A

$$y=2(x + 3)^2 + 7$$

B

$$y=-2(x + 3)^2 + 7$$

C

$$y=-(x + 7)^2 + 3$$

D

$$y=(x + 3)^2 - 7$$

E

$$y=\frac{1}{2}(x - 1)^2 + 9$$

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