CSPE - Parabolas Lesson

Parabolas

In previous courses, you defined a parabola as a quadratic function in terms of x. Those parabolas were concave up or concave down. In this course, we will see parabolas that open up, down, left, and right!

A parabola is the set of all points in a plane that are equidistant from a given point (called the focus) and a given line (called the directrix).

image of a parabola graph on a coordinate plane with vertex, focus, and directrix indicated

Standard Form Equations for Parabolas

$LaTeX: \left(x-h\right)^2=4p\left(y-k\right)$ $\left(x-h\right)^2=4p\left(y-k\right)$		$LaTeX: \left(y-k\right)^2=4p\left(x-h\right)$ $\left(y-k\right)^2=4p\left(x-h\right)$
p>0	p<0	p>0	p<0
Opens Vertically		Opens Horizontally
Vertex: (h, k)		Vertex: (h,k)
Focus: (h, k + p)		Focus: (h + p, k)
Axis of Symmetry: x = h		Axis of Symmetry: y = k
Directrix: y = k - p		Directrix: x = h - p

Notice that the 4p expression is always with the side not being squared!

Watch this video to learn how to graph parabolas.

Graph Parabola Practice

Now that you feel comfortable graphing parabolas, let's practice writing equations. Watch this video to get the basics.

Let's try another problem together.

Write the equation of the parabola shown in the figure.

image of a parabola opening up on a coordinate plane with focus and vertex indicated as follows:
F (3, 2)
y=-1 1. We know that the parabola opens up - so our equation should look like: equation image indicator $LaTeX: \left(x-h\right)^2=4p\left(y-k\right)$ $\left(x-h\right)^2=4p\left(y-k\right)$

2. Our vertex must be halfway between the focus and directrix, which are labeled. So the vertex must be equation image indicator $LaTeX: \left(3,\:\frac{1}{2}\right)$ $\left(3,\:\frac{1}{2}\right)$

3. The distance from the vertex to the focus is equation image indicator $LaTeX: \frac{3}{2}$ $\frac{3}{2}$ so that is p. Let's put all of this into our equation: $LaTeX: \left(x-3\right)^2=4\left(\frac{3}{2}\right)\left(y-\frac{1}{2}\right)\\ \left(x-3\right)^2=6\left(y-\frac{1}{2}\right)$ $\left(x-3\right)^2=4\left(\frac{3}{2}\right)\left(y-\frac{1}{2}\right)\\ \left(x-3\right)^2=6\left(y-\frac{1}{2}\right)$

Just like we converted circle equations from general form to standard form, we can also convert parabola equations! Let's try one: equation image indicator LaTeX: -y^2+4x-8y-44=0 $-y^2+4x-8y-44=0$

Step 1	Let's move y's to one side and x's to the other	$4x-44=y^2+8y$
Step 2	Only one side is squared, so we only need to complete the square for the y's.	$LaTeX: 4x-44=y^2+8y+\dots\\ 4x-44+16=y^2+8y+16\\ 4x-28=\left(y+4\right)^2$ $4x-44=y^2+8y+\dots\\ 4x-44+16=y^2+8y+16\\ 4x-28=\left(y+4\right)^2$
Step 3	Now, we want to factor both sides to reveal the vertex and the value of 4p. Make sure the squared expression is on the left side of the equation.	$LaTeX: \left(y+4\right)^2=4\left(x-7\right)$ $\left(y+4\right)^2=4\left(x-7\right)$

Step 1

Let's move y's to one side and x's to the other

equation image indicator LaTeX: 4x-44=y^2+8y $4x-44=y^2+8y$

Step 2

Only one side is squared, so we only need to complete the square for the y's.

equation image indicator $LaTeX: 4x-44=y^2+8y+\dots\\ 4x-44+16=y^2+8y+16\\ 4x-28=\left(y+4\right)^2$ $4x-44=y^2+8y+\dots\\ 4x-44+16=y^2+8y+16\\ 4x-28=\left(y+4\right)^2$

Step 3

Now, we want to factor both sides to reveal the vertex and the value of 4p. Make sure the squared expression is on the left side of the equation.

equation image indicator $LaTeX: \left(y+4\right)^2=4\left(x-7\right)$ $\left(y+4\right)^2=4\left(x-7\right)$

IMAGES CREATED BY GAVS