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I'm looking for an equation that can help me determine the length of the minor axis.

I know the length of the major axis and have the Cartesian coordinates of a point somewhere on the ellipse.

How can I use these to get the length of the minor axis?

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Does the major and minor axis lies respectively on the x and y-axis? –  Américo Tavares Apr 9 '11 at 12:04
Actually it's the reverse. The major axis is on the y-axis and the minor axis is on the x-axis. The center is 0,0. –  user9348 Apr 9 '11 at 12:06
Then you can write the equation of the ellipse, and you also know that the coordinates of the given point satisfies that equation. –  Américo Tavares Apr 9 '11 at 12:15
Could you give me an example? I'm ashamed to admit that my Math is rather rusty. –  user9348 Apr 9 '11 at 12:22
please add the information that "The major axis is on the y-axis and the minor axis is on the x-axis" to the question. –  Américo Tavares Apr 9 '11 at 14:01
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2 Answers

up vote 2 down vote accepted

Added: In a comment OP states that "The major axis is on the y-axis and the minor axis is on the x-axis."

The equation of an ellipse whose major and minor axis are respectively on the $y$ and $x$-axis is

$$\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1,\qquad (\ast )$$

where $a$ is the semimajor axe and $b$ is the semiminor axe. You are given $% 2a$ and you need to find $2b$. Let the coordinates of the given point be $% (x_{1},y_{1})$. Since it is on the ellipse, its coordinates must satisfy $% (\ast )$

$$\frac{x_{1}^{2}}{b^{2}}+\frac{y_{1}^{2}}{a^{2}}=1.\qquad (\ast \ast )$$

Clearing denominators and then dividing by $y_{1}^{2}-a^{2}$ we get

$$a^{2}x_{1}^{2}+b^{2}y_{1}^{2}=a^{2}b^{2}\Leftrightarrow \left( y_{1}^{2}-a^{2}\right) b^{2}=-a^{2}x_{1}^{2}\Leftrightarrow b^{2}=-\frac{% a^{2}x_{1}^{2}}{y_{1}^{2}-a^{2}}=\frac{a^{2}x_{1}^{2}}{a^{2}-y_{1}^{2}}.$$

Since $a^{2}-y_{1}^{2}\geq 0$ and $b>0$, we obtain

$$b=\frac{a|x_{1}|}{\sqrt{a^{2}-y_{1}^{2}}}.\qquad (\ast \ast \ast )$$

The length of the minor axe is $2b$.

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Thanks, that did the trick. I wasn't sure what the "|" symbols around x1 were supposed to represent so I just wrote it as a * x1 and it worked. The point is now consistently on the ellipse regardless of changes to its position. –  user9348 Apr 9 '11 at 15:32
@Joe: $|x_1|$ is the absolute value of $x_1$ (en.wikipedia.org/wiki/Absolute_value). You need to get a positive value for $b$. –  Américo Tavares Apr 9 '11 at 17:28
Ah yes. That cleared thing up. The days of my schooling are far behind me so it's no wonder I forgot some of the rarer mathematical symbols. –  user9348 Apr 9 '11 at 20:47
Could you please elaborate on how you divide by $x^2 - a^2$? I tried long division and I got something else. For example $a^2 x^2/(x^2-a^2)$ is what? –  Pickett Feb 5 at 9:23
@Calle Thanks! It's a typo, it should be "dividing by $y_{1}^{2}-a^{2}$": \begin{eqnarray*} \frac{x_{1}^{2}}{b^{2}}+\frac{y_{1}^{2}}{a^{2}} &=&1\Leftrightarrow a^{2}b^{2}\left( \frac{x_{1}^{2}}{b^{2}}+\frac{y_{1}^{2}}{a^{2}}\right) =a^{2}b^{2} \\ &\Leftrightarrow &a^{2}x_{1}^{2}+b^{2}y_{1}^{2}=a^{2}b^{2} \\ &\Leftrightarrow &\left( y_{1}^{2}-a^{2}\right) b^{2}=-a^{2}x_{1}^{2} \\ &\Leftrightarrow &b^{2}=-\frac{a^{2}x_{1}^{2}}{y_{1}^{2}-a^{2}}=\frac{% a^{2}x_{1}^{2}}{a^{2}-y_{1}^{2}}. \end{eqnarray*} Corrected. –  Américo Tavares Feb 5 at 18:21
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So you know the length of the semimajor axis, and it's along y. Let's call this axis 'a'. We'll call the length of the semiminor axis 'b'.

x^2 / b^2 + y^2 / a^2 = 1. You also have another point (x1, y1). Simply sub this into the equation and solve for b!

x1^2 / b^2 + y1^2 / a^2   = 1
(a^2 - y^1)/ a^2          = x1^2 / b^2
x1^2 * a^2 / (a^2 - y1^2)`= b^2

Of course, this approach won't work if a^2 = y1^2 (as you'll be dividing by 0), but a point on the ellipse should mean this will never be the case.

I may have made an algebraic mistake somewhere there, but the approach should still be good. :)

Hope this helps.

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'a' is the semimajor axis (and 'b' the semiminor axis). –  Américo Tavares Apr 9 '11 at 13:21
Yes, of course. Sorry for that - my brain usually melts at this time of night! Corrected in my post. –  Ben Stott Apr 9 '11 at 13:26
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