## How Do I Find the Focus of a Parabola?

This is a common question that we get from two audiences. It is most commonly asked by algebra students who want to calculate the point in the plane, when the equation of the parabola is given. More advanced students sometimes must prove this formula from the definition of the focus. We will handle this scenario second. We have also included the process for finding the focus of a parabola when we’re only given the vertex and a point on the parabola.

#### Formula for the Focus

We will assume you are given the vertex form of the parabola. Say you have y = 3(x-1)² + 2. We are asked to find the focus. How do we go about it?

The formula we need in this case is F(h, k+p). Recall that the vertex is V(h,k) when y = a(x-h)²+k. Because we have the equation in vertex form already, we know the vertex is at V(1,2). The only think we need now is p, where p = 1/(4a). As a = 3, in this case, p = 1/(4*3) = 1/12. So F(h,k+p) = (1, 2+1/12) = (1, 25/12). And so, the focus is the point F(1, 25/12).

Thus, to find the focus of the parabola, we (1) find the vertex, V(h,k); then (2) compute p = 1/(4a); then (3) compute F(h,k+p).

Here’s another example.

Example 1: Find the focus of the parabola y = -2(x+4)²-1.

Solution: Because y = -2(x+4)²-1, the vertex is the point V(-4,-1). Next we calculate p = 1/(4a), noting that a = -2, in this case. When we do we find p = -1/8. Finally the focus is the point (-4, -1 + -1/8) = (-4, -9/8).   ■

Sometimes we are given the equation of the parabola in standard form and must either convert the equation to vertex form or use formulas to compute the vertex.

Example 2: Find the focus of the parabola y = (1/2)x² – 5.

Solution: Note that, in this example b = 0, and that any time that b = 0, the standard form and vertex form of the equation are identical. But in this case, we will compute the vertex using the formulas we would use if the vertex form of the equation were not also given to us.

The vertex formula gives us h = -b/(2a) and k = (4ac-b²)/(4a). In this case, a = 1/2, b = 0, and c = -5. Thus h = 0 and k = -5 and the vertex is the point (0, -5). With a = 1/2, then p = 1/2, so that F is the point (0, -5+1/2) = (0, -9/2).   ■

#### Deriving the Formula for the Focus

For more advanced students who wish to prove the formula for the focus, here is an outline of one way to do it.

HOW TO DERIVE THE FORMULA FOR THE FOCUS: Start with the definition of a parabola. A parabola is the set of all points equidistant from a point F, called the focus to a point on a line, called the directrix. Let F be the point (m,n), and the directrix be the line y = t. Now pick an arbitrary point (x,y). We use the fact that his point must be equidistant from F and line D, such that the distance formula from F to P must be the same as P to D. When we equate the distance, we look for a relationship between x and y. What we find is that y is quadratic in x.

Next, because we know y is quadratic in x, write the coefficients of this quadratic as A, B, and C, and set out to write m, n, and t in terms of A, B and C. We encounter a system of three equations in three unknowns. Once we solve each m, n and t, in terms of A, B, and C, we obtain the usual relationships by defining a point V(h,k), called the vertex, where h = -B/(2A), k = (4AC-B²)/(4A). Then if p = 1/(4A), then focus, F(m,n) is given by m = h; n = k+p, and directrix is given by y = k-p.

#### Finding the Focus, Given the Vertex and Point on the Parabola

When we’re given the vertex and a point on the parabola, and asked to find the focus, we’re going to want to use a the following process. First, we write the equation of the parabola in vertex form, to the extent we can. We will be able to write the entire vertex form of the equation except for our leading coefficient, a. Next, we’ll use the point that we’re given as a solution to this equation. This means we can substitute the point (x,y) in for the x and y values in the vertex equation. Then we solve for a. After we have a, we can compute p = 1/(4a), and F(h,k+p), which will give us our focus.

Here’s an example of how we go about it.

Example: Given the vertex is at V(1,2) and the point (3,4) is on the parabola, find the coordinates of the focus.

Solution: First use the vertex to write the equation of the parabola in vertex form to the fullest extent we can. Recall vertex form of the parabola is y = a(x-h)²+k. Because we are given the vertex is V(1,2), the vertex form of the equation of our parabola is y = a(x-1)² + 2.

Next, we use the fact that the point (3,4) is a solution to this equation to find the value of a. We do this by plugging the values x=3 and y=4 into our equation: 4 = a(3-1)²+2. Then we solve to find that a=1/2.

Because we now have the value of a, we can easily find p=1/(4a)=1/2. So, F(h,k+p) = F(1,2+1/2)=F(1,5/2), and our focus is the point F(1,5/2).   ■

• Glenn

I’m confused about 4a and 1/4a. Like some of the questions I dis had the answer 4 multilingual ba. And some of the others had it like 1/4a. How do I know when to use which one.

• http://www.protutorcompany.com/ The ProTutor Company

Hi Glenn — Can you give an example of what you mean?

• Dawn McCullough

If given a problem like (y-2)^=8(x+1), I know the vertex is (-1,2) but
how do I go about finding the focus and the directrix? I know part of
the focus contains the 2, like (a,2). How do I find a? and the directrix
from here? 😉

• http://www.protutorcompany.com/ The ProTutor Company

Hi Dawn — If you already have the vertex, the key to finding the focus and directrix is to find the value of p. Once you have p, then you just add and subtract it from the vertex to find the focus and directrix. In your example, we are given the equation of the parabola in the form, (y-2)² = 8(x+1). Since this is of the form y² = something, we know 8 = 4p, or p = 2. This is because the general version of this form of the horizontal parabola is y² = 4px (for the parent function), and (y-k)² = 4p(x-h) (in the translated case).

Because p > 0, we know that the parabola opens to the right. So we add p = 2, to the x-coordinate of the vertex to get the focus. So the focus will have the x-coordinate -1 + 2 = 1. The focus coordinates, therefore, will be (1,2).

The directrix will then be a vertical line, to the left of the vertex: p units to the left, to be exact. So in this case, the directrix will be the line x = -1 – 2, or x = -3.

To check, we can use the focus and directrix we just found, and generate a couple of points that must lie on the parabola determined by these. Then we can test that those points we generated are indeed solutions to (y-2)² = 8(x+1). Alternatively (and more rigorously), we can just use the directrix and focus we have found to derive the equation of the parabola from the definition. In the case of a directrix of x = -3 and a focus of (1,2), we do indeed find that (y-2)² = 8(x+1). (See the attached photo to see this.)

Thanks for your question. We’re actually presently working on additional materials to help readers with horizontal parabola problems. Your question has helped us to decide some of the topics to hit as we produce new content.

• patrick mburu

Hae Glen i have this question.A dish has the form of a parabola. if the dish is 8ft across its opening and 3ft across its center,at what point should the receiver be placed.

• http://www.protutorcompany.com/ The ProTutor Company

Patrick:

Here is the five step approach I used to solve this one.

Step 1: Make sure you ave a good, accurate diagram, and identify what you need to find.

I’ve attached my diagram of the dish with an 8 ft diameter. By saying that the dish is “3ft across its center,” I interpreted this to mean 3 ft from the plane intersecting the edge of the dish to its lowest point — i.e. the vertex — if we imagine modeling the cross section of the dish with a parabola in the xy-coordinate plane.

Step 2: Use coordinates to find the focus, vertex, and P, a point on the edge of the dish.

If our dish has an 8 ft diameter, this means that it is 4 ft from the axis of symmetry to the edge of the dish.

Now imagine a cross section of the dish in the xy-plane. The cross section makes a parabola. The lowest point in the dish is the vertex. Call that point (0,0).

This means are job to find the placement of the receiver, is to find the point where all reflections converge in the parabola. In other words, we must find the focus of this parabola. When the vertex of the parabola is (0,0), the focus of the parabola is F(0,1/(4a)). This means we just have to find a in order to find the focus, and the location of the receiver.

Finally, if the dish has a depth of 3 ft, this makes the coordinate of an edge point, P(4,3). This is probably the trickiest part of the problem. Once we get the coordinates P(4,3), the rest of the problem isn’t too bad. Take a look at the attached diagram to make sure this makes sense.

Step 3: Use the coordinates of P to find “a” in y = ax².

Now we can model the cross section of the dish with y = ax². If we know the point P(4,3) satisfies this equation, then we can plug this xy-ordered pair in for x and y in this equation: 3 = a*4², and allow us to solve for a. This gives us a = 3/16.

Step 4: Use a to find the focus.

This means that the focus is F(0, 1/(4a)) = F(0, 4/3).

Step 5: Interpret the results within the context of the problem: What does the coordinates of the focus, F(0,4/3) mean with regards to the placement of the receiver?

Finally, we know that because the focus is at F(0,4/3), the receiver should be placed 4/3 ft above the vertex of the parabola, or the lowest part of the dish. 4/3 ft = 1 ft 4 in. So this is the height above the lowest point in the dish where the receiver should be placed.