anonymous
  • anonymous
What is the equation of the ellipse with co-vertices (0, 2), (0, -2) and vertices (3, 0), (-3, 0)? Please help? I have nO IDEA how to do this!
Mathematics
jamiebookeater
  • jamiebookeater
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UnkleRhaukus
  • UnkleRhaukus
hmmm, lets draw a graph, and plot those points
UnkleRhaukus
  • UnkleRhaukus
|dw:1439094439604:dw|
UnkleRhaukus
  • UnkleRhaukus
|dw:1439094488079:dw| can you do the others?

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UnkleRhaukus
  • UnkleRhaukus
(click on the yellow pencil icon, to reuse a drawing )
anonymous
  • anonymous
|dw:1439094586269:dw| is this correct?
UnkleRhaukus
  • UnkleRhaukus
whoops, i plotted that first point wrong,
UnkleRhaukus
  • UnkleRhaukus
(my mistake) |dw:1439094795562:dw|
UnkleRhaukus
  • UnkleRhaukus
[that's better now (sorry about that)] now we have the co-vertices plotted, we can draw the ellipse , so that is goes through these points . . .
UnkleRhaukus
  • UnkleRhaukus
(can you draw the ellipse?)
anonymous
  • anonymous
|dw:1439094980170:dw|
UnkleRhaukus
  • UnkleRhaukus
Good. |dw:1439095016485:dw|
UnkleRhaukus
  • UnkleRhaukus
Can you tell from the diagram now, where the centre of ellipse is ? and its semi-major axes?
anonymous
  • anonymous
(0,0)
UnkleRhaukus
  • UnkleRhaukus
good, yes that is the centre
UnkleRhaukus
  • UnkleRhaukus
And remember that the axes are half the width (a), and half the height (b)
anonymous
  • anonymous
it's 6, and 4, right?
UnkleRhaukus
  • UnkleRhaukus
the major axes (full width/height) are 6 and 4 we want the semi-major axes ... half these
anonymous
  • anonymous
3, and 2
UnkleRhaukus
  • UnkleRhaukus
you now have \((h,k)=(0,0)\) \(a=3\), \(b =2\) so plug these into \[\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2} = 1\]
UnkleRhaukus
  • UnkleRhaukus
What do you get:
anonymous
  • anonymous
(|dw:1439095772471:dw|
UnkleRhaukus
  • UnkleRhaukus
very good, now just simplify
anonymous
  • anonymous
|dw:1439095814555:dw|
anonymous
  • anonymous
y^2
UnkleRhaukus
  • UnkleRhaukus
|dw:1439095846244:dw|
UnkleRhaukus
  • UnkleRhaukus
Great work! you're done
anonymous
  • anonymous
Thank you once again! :)

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