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The Ellipse - Problem 1 9,597 views
Use the height of the ellipse to find the "a" and "b" values and plug into the general equation of an ellipse (x^2)/(a^2) + (y^2)/(b^2) = 1. The "a" value is the horizontal distance from the center of the ellipse to the endpoint (or half the width of the entire ellipse) and the "b" value is the vertical distance from the center of the ellipse to the endpoint (or half the height of the entire ellipse). To find the foci, use the formula sqrt (a^2 - b^2) = c, where c is the distance from the center to the foci.
So what we're going to look at now is an ellipse where we're told that we have it centered around the origin and we're told information about the width and the height. So this equation centered at the origin 10 wide units and 6 units tall.
So the first thing I always tend to do when I get a problem like this is to just draw a little picture so I can see what I'm working with. It doesn't have to be precise and God knows it's not going to be because I really can't draw ellipses very well, but basically it's 10 units wide which means it's going to 10 units from side to side. So that is going to be the length of the x diameter which we divide that into half so the x radius then is going to be 5 units. It's going to be 5 units from the center to a side.
So we go out 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, in each direction. We're also told that it is 6 units tall, so it's 6 units from the bottom to the top, again with the same logic with the width, we that it's then going to be 3 units from the center to the top and three units from the center to the bottom; 1, 2, 3 1, 2, 3 and so what we end up having is an ellipse going through the 6 points. I'm going to try draw it, you can laugh at me when I fail completely, but basically what we're going to end up with is something that looks like a football, actually it's not horrible I've done a lot worse.
So what we're looking for now is this equation and the key thing we need to know is that it is a horizontal ellipse which means that our major axis is going to be the x direction. So the way I would phrase it then is my x radius is my major radius which is going to be 5 and then my y, the minor axis is going to be this one which is going to be the y.
So what that tells us is we come over here, we don't have any transformations going on, so it's just going to be x² over something plus y² over something is equal to 1 and then all you put down here is the x radius basically. So in this case our x radius is 5, so 5² goes in the bottom I mean here and the y radius in this case the minor radius goes here, so this is going to be 3².
Now a brief note this x radius, y radius definitely not textbook, it's probably frowned upon by a lot of mathematicians, but I use it just because it's a really good way to sort of talk about the problem. The main thing in Math terms remembering your major axis as your longer axis here, your minor axis is the shorter, but really you're relating to an x and a y.
So we found our equation for ellipse. I could square 5² and 3, but for practical purposes we have what we need.
The last thing we are looking for is the focus, the Foci and so basically you need to remember the relationship where your Foci are a² minus b² is equal to c², where a is your major radius, b is your minor radius and c is the distance from the center to the Foci. So a² is just relating to our major which is 5², so this just turns into 25 minus 9 is equal to c², so we end up with c is equal to plus or -4.
So what that tells us is our focus, our four units and our focus are always on the major axis, so we're basically going from the center over 4 units in either direction. So I know from there that my focus is I have one at 4,0 and I have the other one at -4,0.
So basically by sketching a picture it's really easy for us to extract the information for this, plug into our equation and then using the relationship a² minus b² equals c² to find our focus.