Algebra 2/Parabolas

In Terms of Distances
Although we define the parabola as an intersection of a cone and plane parallel to the edge of the cone, this is pretty clearly a difficult definition to work with.



But luckily it turns out that there is a proof (due to Dandeline, which can be found in a nice presentation here: | Dandeline's proof) that a parabola is equivalently defined as the locus of points equally distant between some point and some line.

More precisely, we begin with a point which we call the focus, and a line which we call the directrix. In principle these may be any point and line which are not coincident.

For simplicity we will assume (until indicated otherwise) that the focus has horizontal coordinate equal to 0. This will make calculations simpler. The only thing left unknown about the focus, then, is its vertical coordinate.

We will write the focus as $$F(0,c)$$.

Also for simplicity we assume that the directrix is horizontal. Therefore the directrix always has the form


 * $$y=a$$ where $$a\ne c$$

We also define the vertex to be the point which is equally distant from the focus to the line. This makes the vertex either the minimum point (if the directrix is below the focus) or the maximum point (if the directrix is above).

Suppose the focus is $$F(0,2)$$ and the directrix is $$y=-2$$. Find the vertex.

We will primarily consider parabolas for which the directrix is horizontal. In this case we often say that the resulting parabola "opens vertically".

However, once you understand parabolas which open vertically, it is not very hard to understand ones which open horizontally (and have a directrix which is vertical).

Suppose the focus is $$F(0,0)$$ and the directrix is $$x=1$$. Find the vertex.



In general, a point $$P(x,y)$$ is on the parabola if and only if the distance FP equals the distance from P to the directrix.

Because F(0,c) then the distance FP is given by the distance formula,


 * $$d_f = \sqrt{(x-0)^2+(y-c)^2}$$

Because the directrix is horizontal, it is easy to find the distance to P.


 * $$d_d = |y-a|$$

By definition of a parabola, being points at which these two distances are equal, then


 * $$d_f = d_d$$

We define p to be the distance from the vertex to the focus and therefore, trivially, it is always half the distance from the focus to the directrix. It can always be computed by


 * $$p = |c-a|/2$$

If we write the coordinates of the vertex as $$V(h,k)$$ then clearly $$h=0$$ because the focus, too, has its horizontal coordinate at 0. Also, if we assume $$c > a$$ then


 * $$ k = c-p = a+p$$

If on the other hand $$c < a$$ then $$k=c+p=a-p$$.

From the setup described above, derive the equation
 * $$ 4p(y-k) = x^2$$.

Hint: Write out the equation $$d_f=d_d$$ using what you know about each of these. Then square both sides to get rid of the square-root. Then distribute out all operations until you're able to solve for $$x^2$$.

Suppose the focus is at $$F(0,2)$$ and the directrix is $$y=-2$$.

Find all other information. The full set of information in any parabola problem is: the values of $$a,c,p,h,k$$, the coordinates of the vertex, the equation of the directrix, and the equation of the parabola.

Let's consider what happens when the focus does not have its horizontal coordinate at 0. (It will turn out: not much.)

Suppose the focus is at $$F(-2,10)$$ and the directrix at $$y=4$$. Find the equation of the parabola determined by these.

The General Equation: Opens Upward
The above is a demonstration that the equation for a parabola which opens upward is
 * $$ 4p(y-k)=(x-h)^2 $$

As before, the equations $$k=\frac{a+c}{2}$$ and $$p = \frac{c-a}2$$.

a is the position of the directrix, $$F(h,c)$$ the coordinates of the focus, $$V(h,k)$$ the coordinates of the vertex, and p the distance from vertex to focus.

In this exercise we will see that if we know the equation of a parabola, then we can recover all of the information: $$a,c,p,h,k$$.

This means that if you know how to find the equation, then you know how to find everything.

Suppose a parabola has equation $$\frac 1 4(y+1) = (x-1)^2$$. Find the parameters $$a,c,p,h,k$$.

The General Equations: Downward, Rightward, Leftward
Until now we've assumed the directrix is horizontal, and the focus is above the directrix. This causes the parabola to open upward.

If the focus is below the directrix, all the same analysis applies. You will find that the only thing which changes is


 * $$p = -\frac{c-a}2$$

which is necessary so that p is positive. This is necessary because p is a distance and so must be positive.

Then downstream of that, we get an equation which is only slightly different.
 * $$ -4p(y-k)=(x-h)^2$$

Suppose the vertex of a parabola is at $$V(1,2)$$ and the focus is at $$F(1,0)$$. Determine the equation of the parabola.

Next consider if the directrix is vertical rather than horizontal. Now the vertex and focus will share the same vertical coordinate by may have different horizontal coordinates. We now say that $$F(b,k)$$ and $$V(h,k)$$ are the coordinates of the focus and vertex, and as before p is the distance from focus to vertex.

Much like before,
 * $$h = \frac{a+b}2$$

If the focus is to the right of the directrix, the equations become


 * $$p = \frac{b-a}{2}$$

and the equation becomes


 * $$4p(x-h)=(y-k)^2$$

If the focus is left of the directrix,


 * $$ p = -\frac{b-a}2$$

and the equation becomes


 * $$-4p(x-h)=(y-k)^2$$

A consolidated display of the four cases is shown below.

A satellite dish is in the shape of a parabola, as shown.

It turns out that, if a light ray comes into the dish and strikes the surface, it will always reflect off of the surface and be directed into the focus. This is shown on the left.

Assume that the rim of the dish has a diameter of 20 feet. Also assume that the depth of the dish is 10 feet.

Find the point at where all of the incoming rays will intersect.



