Find the standard form of the equation of an ellipse with vertice...

Question

Find the standard form of the equation of an ellipse with vertices at (0, -6) and (0, 6), passing through (2, 4).

Accepted Answer

Hello everyone in this video. We're going to be looking at this practice problem where we want to write the standard form of an ellipse equation given that the ellipse has vortices at zero negative nine and zero positive nine and it passes through the 90.12 and three. In order to solve for this problem. Let's recall the standard form of and a lips. So remember that the standard form of an ellipse follows x minus H squared divided by a squared plus y minus K squared divided by b squared is equal to one. And also recall that the center of the lips is at the positions H. K. So that's where these values come from. So we're going to need to solve for each of these values to figure out the standard form for our ellipse. So for our lips let's try to visualize where the center would be at. So say a draw. An ellipse were given the vertex is for our lips. And recall that the vertex is are on the major access and they're a units from the center. So the vertex is are going to be key in us finding the center. So this top vortices at 09 and the slower vertex is at 0 -9. And notice when I say the major axis, I mean the axis of cemetery for this ellipse. And notice how it runs vertically which means it runs parallel to the Y axis. And for our center we want to find this cross or this point here. So because are vortices running along this access? Notice how the center is going to have the same X values as our furnaces. And that makes sense because our burgesses actually follow the form H. K plus or minus A. So I know that my center has a coordinate of zero for the X. And now I need to find the midpoint between both of my versaces. So this is the total length for between my versus is I need to find this midpoint. And because from this graph you can see that I go nine units below this axis, this X axis I get my first versus E. And then if I go nine units above, I get my upper vertex. The center would actually be at zero. And we can figure that out using the midpoint formula where I'll have my first For a text - plus my second brew text positive nine divided by two. And this becomes negative nine plus nine is 0/2. So this is actually zero. So I know that my Y axis or my Y coordinate for my center is zero. And if you can tell 00 is at the origin. But nonetheless we got the values for H and K where H is equal to zero and K is equal to zero. And if I plug those into my standard equation, I'll have x minus zero squared divided by a squared plus y minus zero squared divided by b squared is equal to one. So that simplifies to x squared divided by a squared plus y squared divided by b squared is equal to one. And now we can solve for are next values which are A. And B. And we can actually only solve for A. Because if we go back to the definition for the coordinates of adversity, you can see that we have a right here. So I can only solve for this A value. And because this is a standard form it seems like we can just start plugging in but notice how I said that the major axis was the Y axis. That means that we need to switch are A squared and R. B squared because the major axis is the Y axis. We need to make sure that this a variable is associated with the Y value. So our standard form equation is actually X squared divided by B squared plus y squared divided by a squared is equal to one and now my A. Is associated with my Y. But let's go ahead and try to find the value of A. So if I look at my lower vertex, C. Or lower vertex It's at 0 -9. And that is equal to the H. K minus a value. So I already know that H is equal to zero. But now I have k minus A. Is equal to negative nine. And I know that K is zero. So this becomes zero minus A. Is equal to negative nine or negative A. Is equal to negative nine. If I get rid of that Negative I have a. Is equal to positive nine. And I can go ahead and test this with my upper vertex. So my upper vertex is at zero and positive nine. That is equal to H. K. Plus A. Notice how it's A. Plus A. Now already know that H. Is equal to zero and now I have K. Plus A. Is equal to positive nine and K. Is equal to zero. So I have A. is equal to nine. So I got the same value of a. For both of my vertex or versus is. So I'm gonna go ahead and plug that into my standard equation. So now I have X squared divided by B squared plus y squared divided by nine squared is equal to one or x squared divided by B squared plus y squared divided by 81 Because nine times 9 is 81 is equal to one. And now I still need to solve for this B value but I don't really have any other definitions that I can use. So this is where the point that they gave us 12 and three is going to come into play. Because if I plug in A Y. Value in an X. Value, the only variable left to solve is the B. So I'm gonna plug in V 12 and three coordinate into my standard equation to get a value for B. So if I do that, I have 12 squared divided by B squared plus three squared divided by Is equal to one. So 12 times 12 is 144 divided by B squared plus three times three, which is nine divided by 81 is equal to one. I'm going to go ahead and simplify 9/81 to 1/9. And from here I can move over the 1/ to the right hand side. So now I have divided by B. Squared, is equal to 1 -1 9th. And if I rewrite the one as 9/9, I have 1 44 divided by B squared is equal to 8/9. And because the B squared is in the denominator, I'm going to multiply by B squared. So I'll have 144 is equal to eight divided by nine times B squared. And in order to move this fraction over, I'm going to multiply by 9/8 to both sides. So I'm left with nine times 44 is 1296, divided by eight. That is equal to B. 1,296, divided by eight is 1 62. And because I have the B squared, I'm going to take the square root of both sides And 162 is the same thing as 81 times two. So I'm going to go ahead and separate my square root. I'm actually going to rearrange this, so B is in front, so I'll have B is equal to the square root of 81 times the square root of two. And notice here how the square root of 81 is nine. So I can say B is equal to nine times the square root of two. And now I have a value for B and I can go ahead and plug that back into my standard equation here. So be is nine times Radical two. So now you can see that I have a complete standard equation where this is X squared divided by nine times square root of two squared plus Y squared divided by nine squared is equal to one. So this becomes my final answer. And if we look at the answer choices, you can see that this aligns with answer choice B. So hopefully this video hoped and see you next time.

Find the standard form of the equation of an ellipse with vertices at (0, -6) and (0, 6), passing through (2, 4).

Key Concepts

Standard Form of an Ellipse

Vertices of an Ellipse

Substituting Points into the Equation

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