Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

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Focus and Directrix of a Parabola - Problem 2

Carl Horowitz
Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

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Finding the focus and directrix given a quadratic equation. So what we have behind me is a quadratic equation, it's not quite in the form we're used to, but it still has a x² and a constant term and a single y term and we're trying to find the focus and directrix.

The first thing we have to do when looking for the focus and directrix is to find the vertex of our parabola and we have two different ways of doing that. We have to get y equals first and foremost and we could either complete the square, or we can just do -b over 2a to find the x coordinate of the vertex, either one is perfectly fine.

I'm going to go ahead and just do -b over 2a, but if you want to complete the square, you could do that as well. So the main thing we have to do is isolate our y by itself, so what I'm going to do is add the 4y over to the other side giving us x² minus 4x plus 16 is equal to 4y and then just divide by 4 giving us 1/4 x² minus x plus 4 is equal to y.

So we now want to find the vertex. X coordinate and the vertex is just -b over 2a, negative -1 is 1, 2 times 1/4 is 1/2 and so one over 1/2 is just going to equal 2. So we found that the x coordinate of the vertex is 2, we now need to find the y coordinate of the vertex, just plug in 2, so what we end up with is 2² is 4, 1/4 times 4 is 1, 1 minus 2 is -1 plus 4 is 3.

So what we are able to do is find the vertex. So what I actually have here is a parabola with my center over two units and up three units facing upwards, very rough sketch. So I know that my focus has to have the x coordinate the same as the vertex because the focus is just directly in line on that axis of symmetry and my directrix is going to be below it. So what I need to do is find the distance between the focus and the vertex, and the way we can do that is by using our formula, absolute value of a is equal to 1 over 4c where c is that distance between the focus and the vertex.

'A' is just a coefficient from our quadratic which we know to be 1/4, so we have 1/4 is equal to 1 over 4c telling us that c is equal to 1. So we know that c is 1, c is the distance between the vertex and the focus, so to find the focus, all we have to do is go up one unit and I know I have to go up one unit because this is an upward facing parabola, that focus has to be in the center.

So we know that our focus, we know the x coordinate is the same 2 and then our y coordinate is up one which is 4, and then our directrix is going to be a horizontal line remember it's going to have to be somewhere down here and it's exactly the same amount of measure below the vertex as the focus is below it. So we went up one unit to find our focus, we go down one unit to find our directrix, vertex y value is 3, we go down one so we end up with y is equal to 2 for our directrix.

So whenever we're looking for information regarding a quadratic related to the focus and directrix, the main thing we want to do is find our vertex and then we can use this relationship absolute value of a is equal to 1 over 4c to find that distance between the vertex, focus and the directrix.

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