# Focus and Directrix of a Parabola - Problem 1

Finding the equation for a parabola when we have the equation about the focus and the directrix. What we're looking at in this problem is a parabola with a focus at 0,3 and the directrix at y equals -3 and we are trying to find the equation for this parabola.

The first thing I find easy is to draw a little picture so I can see what exactly is going on. So I have my graph, I know my directrix is the horizontal line y equals -3 and I know that my focus is the point 0,3.

From here I can find my vertex because I know that my vertex is directly between my focus and my directrix, so if my focus is 3 units up and my directrix is 3 units down, my vertex then is at 0, 0.

So what I'm going to do is do this two different ways. First I'm going to do just using the distance formula and the second one I'm going to do it using the formula that we know relates to the equation for the graph with the distance between the focus and the vertex.

SO the first way we want to do is basically we have a parabola, I know this parabola is facing upwards because the focus is inside the parabola, so I know that we have a parabola that looks something like this roughly. We also have a point on this parabola x, y and how we know the focus and the directrix work is that the line from the focus to that point x, y is going to be equidistant to that line that connects to the directrix.

That line connecting this line to the directrix shortest way is keeping the x value from here and the y value from here, so I know that this point on the directrix is actually x and -3. So we have this point x, y the x coordinate stays the same to get down to here. The y coordinate is from a directrix -3.

So now what we have are these two segments and the distance from those two segments has to be equal, plug them into the distance formula. So distance formula hopefully you remember distance equals to the square root of change of x² plus change of y², so we have to do the distance between these two things, this point here is 0, 3 so what we end with is the square root of x minus 0² plus y minus 3² is equal to the square root of the differences here. So what we end up with is x minus x² those disappear alright x minus x to squared plus the difference of our ys, y minus -3 is going to turn into y plus 3² and we just need to solve out this equation.

So what we have is the square root of both sides, so we can square both sides in order to get rid of that, that will get rid of my square roots. Then all I want to do is to simplify what I'm working with. Foiling out the x minus 0², that's just going to be x² plus Foiling this out y² minus 6y plus 9 is equal to my x² disappear here, y² plus 6y plus 9 and let's see what we're left with.

So what I see is our +9s cancel, our y²s cancel and so what we are left with is I'm going to bring this 6y over so we end up with x² is equal to 12y. We're used to equations being y equals, so divide by 12 and we end up with y is equal to 1 over 12x².

So I said we're going to do this two ways, this way was the more involving we're using the distance formula, the other way we just have to remember our equation relating the coefficient and x² to our distance between our vertex and our focus.

And what that was was obviously the absolute value of a is equal to 1 over 4c where c is the distance between the focus and the vertex, so we know that our distance between our focus and our vertex is just 3 because we are dealing with a vertex and the point 0,3 so all we have to do then is plug in 3 for that and we end up with the absolute value of a is equal to 12.

We know that the vertex of this parabola is at the origin, so therefore we don't have any horizontal or vertical shifts, so all we have is y is equal to 1/12x² and I know this is a positive one-twelfth because we have an upward facing parabola, therefore our coefficient has to be positive.

Two ways of finding the equation giving the information about the focus and the directrix. Obviously using this formula is a little bit easier, a lot less steps, but if you forget it you can always go back to the fact that any point on your curve is going to be equidistant from your focus and your directrix.

## Comments (0)

Please Sign in or Sign up to add your comment.

## ·

Delete