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The Ellipse - Problem 3 6,690 views
Finding the equation for an ellipse when you're given information about the foci and the co-vertice. So if you remember your co-vertice is actually the end points of you minor axis, so vertices are the end point of your major axis, your longer axis, co-vertices are the end point of your minor or your shorter axis.
So what we want to do first is just sort of plot these points so we can see what we're working with. So I know that my foci are down the x axis six units let's use a different color, so I know that my foci are going to be here and over here and my co-vertices are up and down 8 on the y axis two, four, six, eight, two, four, six, eight.
So what we know is that our foci always lie on the major axis which tells me that my vertices, the end points of my major axis are actually going to be our here somewhere. So we know that it's going to be a horizontal ellipse because it's going to be wider than it is tall. The thing that we have to find now is what that major axis, what our x radius is actually going to be.
I know that my center has to be at the origin because the only way for our minor axis and our foci to be on perpendicular lines which our axis are, is for them to cross right at the origin, so therefore I know I don't have any transformations going on. So from that I can easily write that the equation for this is going to be x² over something plus y² over something is equal to 1. The y is over it's minor radius again it's a horizontal so the y is going to be over the y radius squared, y radius is the distance from the origin to the top of that axis, so this is just going to be over 8² or 64.
We now need to find out what this measurement is and we can do that using a relationship with a² minus b² is equal to c². A² is the major radius which is going to be the distance from the vertex to whatever that point is, that vertice is we don't know what it is. B² is the minor radius which we do know, which is just going to be this, so that's just going to be a² minus oops that's 8 isn't it? So yeah it's 8 up here so that's just going to be what I said minus 64 I'm confusing myself up here is equal to c² which is the distance from the center to the foci which is just going to be 6, so this is just going to equal 36.
Solving for a, add 64 over a² is equal to 100 which tells us that a is equal to 10. 10 then is going to be our major radius. So we're going to be center which at the origin out 10 units to find that vertice that end point of the major axis, so 10 is what we squared to put into our equation, so really all we have to do to finish this up is take this and we know we then have 10² underneath our x² term.
So interpreting our information when we're given foci and co-vertices, it be pretty similar if we were told the vertices as well except instead of plugging in for our b, we will just be plugging in for our a.