A parabola is the familiar shape seen in many physical applications, like the path taken by a ball thrown upwards. This lesson explores equations for the parabola and shows how they may be obtained from two quantities: the focus and the directrix.
Remember the last time you played with blocks? Great fun! With lettered blocks, you can build words. It is amazing so much variety exists with only 26 letters. Also, the shapes of letters themselves are inspirational. For example, a parabola is shaped like the letter U.
The Letter U
In this lesson, learn to build various equations for the parabola given only the information contained in the focus and the directrix. Now those are words we probably did not see with our blocks.
Start by drawing a line on a graph. In this example, the line goes through y = -3 and is parallel to the x-axis. This line is called the directrix. Locate a point. In this example, the point is at (3,1). This point is called the focus.
The focus and directrix
The idea is to use only this information to plot a parabola and to find equations that define the parabola. The parabola corresponding to this directrix and focus looks like this:
The focus, directrix, and parabola
The distance from the focus to a point on the parabola curve is the same as the shortest distance from this point on the parabola curve to the directrix line. So, the points on the parabola curve are equidistant (equal distance) to the focus and to the directrix. Finding all the equidistant points is one way to plot the parabola. The blue lines on the image below show the equal lengths.
The turning point on the parabola is the vertex. On this example, the vertex is at (3, -1), which we observe to be halfway between the focus and the directrix. Lots of words to build with our blocks.
Instead of building words, this is building equations. Say the focus is here: (xo,yo). The point on the line is (x,y). The location on the directrix will be at (x,α). Let's keep things general on purpose. In addition to the variables x and y, the equations will have your three numbers xo, yo and α.
Parabola Equation in Standard Form
The parabola equation in standard form for finding the length between two points is the square root of the sum of the squared differences in each coordinate. For this case, the length of the line segment between the focus at (xo,yo) and the point on the curve (x,y) is the square root of (xo - x)^2 + (yo - y)^2. The length of the line segment between the point on the curve (x,y) and the point on the directrix (x,α) is the square root of (x - x)^2 + (y - α)^2. The lengths are equal so you can equate the two square roots. Then, square both sides, expand, and isolate the y on the left-hand side of the equation.
The right-hand side depends on xo, yo, and α. This equation has the same form as y = ax^2 + bx + c. This is the standard form equation for the parabola. Do you see how to find 'a', 'b', and 'c'?
Our example parabola in standard form is y = .125x^2 - .75x + .125.
Parabola Equation in Intercept Form
Sometimes you want to know where the parabola intercepts the x-axis. To get an equation with this intercept information in it already, use the standard form equation with 'a', 'b', and 'c'. Then, factor out the 'a' to get y = a(x^2 + bx/a + c/a). Then use the quadratic equation to find the two roots p and q. The equation looks like y = a(x - p)(x - q). Now, you have an equation which directly gives you the intercept information! The 'p' and 'q' are the x-intercepts. These are the places where the parabola curve crosses the x-axis. The example parabola equation in intercept form is y = .125(x - 5.83)(x - .172).
Parabola Equation in Vertex Form
Sometimes you'll want to know where the parabola has its vertex. To get an equation having this vertex information in it already, take the standard form equation and complete the square. The result is y = a(x - h)^2 + k. You now have an equation in terms of the vertex. The 'h' and 'k' are the x and y coordinates of the vertex. This is the parabolic equation in vertex form. From our example, the location of the vertex is at (3, -1), which tells us 'h' = 3 and 'k' = -1.
Thus, the parabolic equation in vertex form for this example is y = .125(x - 3)^2 -1.
What happens if there are no x-intercepts? How could this happen? The parabola's curve might not cross the x-axis. When trying compute the p and q values of the intercept form, you'll be attempting to take the square root of a negative number. This tells us no x-axis intercepts exist.
If the parabola is concave downward, then 'a' will be negative. If the parabola is sideways with it's opening to the right or left, obtain the expressions by interchanging x and y. Building on our knowledge of the parabola is very much like building with blocks. The possibilities are vast.
The directrix and the focus provide enough information to write an equation for a parabola. Any point on the parabola is equidistant to the focus and the directrix. Equating these distances leads to the parabola equation in standard form.
There are other convenient forms for the standard equation depending on what information you need. For example, solving for the roots of the standard equation gives the parabola equation in intercept form. Completing the square in the standard equation gives us the parabola equation in vertex form.