[SOLVED] Equation of an ellipse...

Printable View

May 6th 2006, 03:23 PM
pidgezero_one

[SOLVED] Equation of an ellipse...

I'm so sorry if this doesn't belong here. ><; I'm going a bit frantic across homework help forums looking for responses to this. @_@;

Does anyone know how to calculate the equation of an ellipse (in this case, (x^2/b^2) + (y^2/a^2) = 1 since I graphed the points and it looked like the major axis was vertical) when the only given information is that the center is (0,0) and that 2 points on the ellipse are (2, -2) and (-2/3, 10/3)?

Any help would be greatly appreciated ^^;

EDIT Nevermind I solved it. Sorry 'bout that.
May 6th 2006, 05:08 PM
topsquark

Quote:

Originally Posted by pidgezero_one

I'm so sorry if this doesn't belong here. ><; I'm going a bit frantic across homework help forums looking for responses to this. @_@;

Does anyone know how to calculate the equation of an ellipse (in this case, (x^2/b^2) + (y^2/a^2) = 1 since I graphed the points and it looked like the major axis was vertical) when the only given information is that the center is (0,0) and that 2 points on the ellipse are (2, -2) and (-2/3, 10/3)?

Any help would be greatly appreciated ^^;

EDIT Nevermind I solved it. Sorry 'bout that.

I see that you got it. I'll answer it anyway, for the sake of completeness.

The general form for an ellipse is:
$\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1$ where the center of the ellipse (the geometric center, not one of the foci) is at the coordinates (h,k).

Since we know the center is at (0,0) we know that h = k = 0 and the ellipse equation is:
$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$

Now, we have two points on the ellipse: (2,-2) and (-2/3,10/3). Putting these points into the ellipse equation gives us (respectively)
1. $\frac{4}{b^2}+\frac{4}{a^2}=1$
and
2. $\frac{4}{9b^2}+\frac{100}{9a^2}=1$

So we have two equations for a and b. This should suffice to solve the system. Equation 1 gives us:
$\frac{1}{b^2}=\frac{1}{4}-\frac{1}{a^2}$

Putting this into equation 2 gives us:
$\frac{4}{9} \left ( \frac{1}{4}-\frac{1}{a^2} \right ) +\frac{100}{9a^2}=1$

$1 - \frac{4}{a^2} + \frac{100}{a^2}=9$

$1+ \frac{96}{a^2}=9$

$\frac{a^2}{96}=\frac{1}{8}$

$a = \pm \sqrt{\frac{96}{8}}=\pm \sqrt{12}=\pm 2 \sqrt 3$

Since a is the length of the semi-major or semi-minor axis, we choose the positive value.

This gives a value for b as: $b = \sqrt 6$ .

Thus the equation for the ellipse is:
$\frac{x^2}{6}+\frac{y^2}{12}=1$ .

-Dan
May 6th 2006, 10:36 PM
CaptainBlack

Quote:

Originally Posted by topsquark

I see that you got it. I'll answer it anyway, for the sake of completeness.

The general form for an ellipse is:
$\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1$ where the center of the ellipse (the geometric center, not one of the foci) is at the coordinates (h,k).

This is the general form of the equation of an ellipse with the major axis
aligned with either the x-axis or y-axis. Since we are given that the centre is
at (0,0), we are in general left with three degrees of freedom (eccentricity,
length of semi-major axis and orientation). Now we are given two points,
which gives us two equations and three unknowns, so in general there is no
unique solution.

So in order to solve this we need the additional information/assumption about the orientation.

RonL