Root of whatever is in the denominator, a is the x-radius. Now, a is the length of the radius in the x-direction. What this means, if this was our ellipse in question right I could have written this asĬ squared and d squared. Squared plus y squared over b squared is equal to 1. So the general or the standardįorm for an ellipse centered at the origin is x squared over a Well, with that said, let'sĪctually go a little bit into the math. Other words, we are always the exact same distanceĪway from the origin. The origin is the same distance as the closest we get, or, in Of this, because in a circle's case, the furthest we get from Really far away from the origin and that's about as far as We'll ever get, just as close as well get down here. We're on this point on the ellipse, we're reallyĬlose to the origin. Let me say this isĬentered at the origin. Or the distance from the center is always changing. Of the circle, while in an ellipse, the distance from theĬenter of the circle is always changing. It's a special case because inĪ circle you're always an equal distance away from the center And the circle is really justĪ special case of an ellipse. Hope that helps, i can clarify more if you still dont get the last video, we learned a 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. Same thing if you shift the Y coord of the center down 2: so x=1 or x=-1 and your points on your x axis are To find your 2 x intercepts you want Y to equal zero. (-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. 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. These are now your 2 points on the ellipse on either side of the center point in the X directions. You can use your center point that you put in for y which was -2 and these x's of your center point, it cancels the Y term out of your equation so that you can solve for your x values. If you make the numerator zero, by putting in your Y coord. 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. Yes, after awhile you will notice the general equations for conic sections with practice. 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. x=1 and x=-5 so (1,1) and (-5,1) are 2 major points on the elipse. for y and solve for x to get your 2 x direction points, (x^2+4x+4=9. Ive typed the reasons in previous questions, read all those and you should get it. (-2,1) is the center of the elipse, it will not satisfy the equation because it is not ON the elipse.
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