anonymous
  • anonymous
What is the equation of the following graph?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
UnkleRhaukus
  • UnkleRhaukus
looks like an ellipse, where is it centred? what are the semimajor axes?
anonymous
  • anonymous
it is, and isn't it centered at (0,0)? and I don't know them..

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UnkleRhaukus
  • UnkleRhaukus
yes it is centred at the origin: (0,0)
UnkleRhaukus
  • UnkleRhaukus
Semi major axes of an ellipse look like this |dw:1439092799738:dw|
anonymous
  • anonymous
|dw:1439092817312:dw| we have to use this equation, correct?
UnkleRhaukus
  • UnkleRhaukus
A general form of an ellipse centred at the point \((h,k)\), with semi major axes \(a\), and \(b\), (in the x and y directions respectively ) is \[\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2} = 1\]
UnkleRhaukus
  • UnkleRhaukus
yeah we have the centre \((h,k)\) = \((0,0)\) so it reduces to \[\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1\]
UnkleRhaukus
  • UnkleRhaukus
we just need to get \(a\), (half the width) and \(b\), (half the height)
anonymous
  • anonymous
how do we get that?
UnkleRhaukus
  • UnkleRhaukus
look at the diagram,
UnkleRhaukus
  • UnkleRhaukus
|dw:1439093170591:dw|
anonymous
  • anonymous
so then A would be 1.5, and B would be 3 ?
UnkleRhaukus
  • UnkleRhaukus
not quite, the full width of the ellipse is 3 - (-3) = 6 the full height of the ellipse is 6 - (-6) = 12 so half the width is 6/2 = ... and the half height is 12/2 = ......
anonymous
  • anonymous
3, and 6, or is it in that fraction form?
UnkleRhaukus
  • UnkleRhaukus
thats right the semi-major axis (in the x direction) : a = 3 and the semi-major axis in the y direction: b = 6
UnkleRhaukus
  • UnkleRhaukus
So what does the equation of our ellipse look like now?
anonymous
  • anonymous
|dw:1439093577574:dw|
UnkleRhaukus
  • UnkleRhaukus
goood, now just simplify 3^2 and simplify 6^2
anonymous
  • anonymous
9 & 36
UnkleRhaukus
  • UnkleRhaukus
Cool, so your equation is \[\frac{x^2}9+\frac{y^2}{36} = 1\] A further step to make the final equation a little nicer could be to multiply both sides by 9 \[x^2+\frac{y^2}{4} = 9\] but you might prefer to keep it as \(\frac{x^2}9+\frac{y^2}{36} = 1\)
anonymous
  • anonymous
Thank you! :)
UnkleRhaukus
  • UnkleRhaukus
i suppose another option for a nice final form might be \[\left(\frac x3\right)^2+\left(\frac y6\right)^2=1\] They all kinda look nice.

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