If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## NASA

### Unit 2: Lesson 3

Orbital mechanics

# Intro to ellipses

Sal introduces ellipses and shows how their standard equation relates to their center and radii (ellipses have two radii: major and minor). Created by Sal Khan and NASA.

## Want to join the conversation?

• What if I have an equation that has coefficients? For example:
(25x^2) + (16x^2) = 400
Would a^2 then be equal to 1/25? And what would I do with the 400?
• (I'm assuming your second x is supposed to be a y, otherwise this is just an algebraic equation)
you first divide both sides by 400 to get (25x^2)/400 + (16y^2)/400 = 1
then rearrange that to get (x^2/400/25) + (y^2/400/16) = 1

a^2 = 400/25 = 16
a=4

b^2 = 400/16 = 25
b=5
• In the formula, how can it equal 1 when it has x and y value of 0?
• I was also struggling with that question, but after some thinking it became pretty clear. The ellipse is constructed out of tiny points of combinations of x's and y's. The equation always has to equall 1, which means that if one of these two variables is a 0, the other should be the same length as the radius, thus making the equation complete. Which is exactly what we see in the ellipses in the video.

We find the center by making x and y equal 0, but not at the same time:

This is what happens:

When x=0 then y=1 (and remember that r=1 always!)
x^2+y^2=r^2, so 0 + 1 = 1

When y=0 then x=1
x^2+y^2=r^2, so 1 + 0 = 1

The equation itself doesn't match (0,0) (only if r=0, which is never the case), but the above method gives us a way to search the exact center of the ellipse. It's quite as simple as that. I hope my explanation helps you (and maybe others) to understand the concept behind this equation!
• Why do ellipses and hyperbolas equal one?
• Because they are 'deformations' of a circle. You know what an equation of a circle looks like, right? Suppose you have this one:

(x - 3)² + (y - 2)² = 4

The center is at (3, 2), but how can one squeeze the circle to make it appear as an ellipse or hyperbola? Divide both side by 4 and you get:

(x - 3)²/4 + (y - 2)²/4 = 1

It's still the same circle, but now you know where that 1 is coming from and you can also squeeze it by changing the denominator of either the x or y term. In this video you can see what happens when the denominator of one of the terms changes:

https://www.screenr.com/lBZN

The equation 'd' is the one I've written above and equation 'e' is:

(x - 3)²/4 + (y - 2)²/b = 1

Where b is the variable that we're changing. Notice that when b = 4, it forms the same circle as 'd', but when b =/ 4 and still positive it's an ellipse. When it goes to negative, it becomes a hyperbola.
• @; ((y-1^2)/4)+((x+2^2)/9)=does not eqUAL 1? Am I suppose to know right off the bat that the problem has to do with ellipses (shifting). First of all, when I substituted in (-2,1) for the x and y into the equation, it didn't equal 1. I tried what you're suppose to follow for order of operations, and that gave me 0/4+0/9, which is not equal to zero, I even tried factoring just to see if it would work, and it didn't: 0/4+8/9.
• (-2,1) is the center of the elipse, it will not satisfy the equation because it is not ON the elipse. Ive typed the reasons in previous questions, read all those and you should get it.
Put in the center coord. for y and solve for x to get your 2 x direction points, (x^2+4x+4=9 ... x=1 and x=-5 so (1,1) and (-5,1) are 2 major points on the elipse. Do the same for your X center coord. and solve for Y. OR once you get used to the standard form of the ellipse you can see that a^2=2 and b^2=3, so you can just use that to add and subtract 2 from your y coord of center point and add and subtract 3 from your x coord of center point to get the same information.
Yes, after awhile you will notice the general equations for conic sections with practice. You will get some equation on a test that doesn't look like a conic at first, but you need to do algebraic manipulations like completing squares to put the given information into the standard forms of ellipses, parabolas, hyperbolas...

Hope that helps...
• how can we recognize what conic section it is when the equation is not written in standard form?
• For a parabola: there is only one squared term in the equation
Circle: the coefficents are the same for the two squared terms
Ellipses: the two fractions are added together
Hyperbola: the two fractions are subtracted from each other
• What is the difference between an oval and an ellipse?
• The term "oval" isn't really used much in geometry because it does not have a very clear definition. Thus, you'd have to ask for clarification if someone mentioned an oval in geometry.

Since oval does not have a standard exact meaning in mathematics, we cannot really compare it to an ellipse which does have a clear meaning.
• If you were to shift the ellipse down 2 spaces, why would you add 2 to y? Wouldn't you subtract 2 from y? Why is it like the opposite of what I think? In other words, why would you want the numerator to equal zero? This seems counter-intuitive.
• Think about where your center point is. In this example it is C(5,-2). If you make the numerator zero, by putting in your Y coord. of your center point, it cancels the Y term out of your equation so that you can solve for your x values.
(x-5)^2=9
x^2-10x+25=9
x^2-10x+16=0
(x-2)(x-8)=0 .... so x=8 or x=2
You can use your center point that you put in for y which was -2 and these x's
(8,-2) and (2,-2)
These are now your 2 points on the ellipse on either side of the center point in the X directions.

If you put in (y-2)^2/25 instead: Try to put in your Y center coord now and see what you end up with. (-2-2)^2/25 = (16/25) so your Y term doesnt cancel out and allow you to find the x direction major axis points on the ellipse.

Think about just a circle x^2+y^2 = 1
To find your 2 x intercepts you want Y to equal zero. Put in Y=0 and you have x^2=1 .. so x=1 or x=-1 and your points on your x axis are
(1,0) and (0,1)

Same thing if you shift the Y coord of the center down 2:
x^2+(y+2)^2=1
Here your center is (0,-2) from shifting down 2, and notice you put a (y+2) term in because you want the Y term to cancel by becoming zero when you put the center point's Y term into the equation.

Hope that helps, i can clarify more if you still dont get it

Pantera62345@hotmail.com
• How would we make an equation of a tilted ellipse?