Ellipse standard equation from graph | Precalculus | High School Math | Khan Academy

So we have an ellipse graph right over here. What we're going to try to do is find the equation for this ellipse.

So like always, pause this video and see if you can figure it out on your own. All right, so let's just remind ourselves of the form of an equation of an ellipse.

So let's say our ellipse is centered at the point. I'm going to speak in generalities first, and then we'll think about the specific numbers for this particular ellipse.

So say the center is at the point (H, K), and let's say that you have a horizontal radius. So the radius in the X direction, horizontal radius, is equal to a. And let's say your vertical radius, let's say your vertical radius, is equal to B.

Then the equation of this ellipse is going to be:

((x - h)^2 / a^2 + (y - k)^2 / b^2 = 1).

So what are H and K, and a and b in this situation? Well, H and K are pretty easy to figure out. The center of this ellipse is at the point.

See, the x-coordinate is -4, and the y-coordinate is 3. So this right over here is -4, and this right over here is 3. And what is a going to be?

Well, a is your horizontal radius, your radius in the horizontal direction. So it's the length of this line right over here, and we can see it's 1, 2, 3, four, five units long. So a in this case is equal to 5.

So this is going to be (5^2), and B is our radius in the vertical direction. We can see it's 1, 2, 3, 4 units, so B is equal to 4.

So that is 4. So we can rewrite this as we could rewrite this as:

((x - (-4))^2 / 5^2 + (y - 3)^2 / 4^2 = 1).

Yus, the y-coordinate of our center.

So (y - 3^2) over our vertical radius squared, so (B^2) is going to be 16, and that is going to be equal to 1.

And of course, we could simplify this a little bit. If I subtract a negative, that's the same thing as adding a positive. So I can get rid of I can just, instead of saying (x - (-4)), I could just say (x + 4).

And there you have it! We have the equation for this ellipse.