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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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    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.2011-04-09

2 Answers 2

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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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    @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.2014-02-05
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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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    Yes, of course. Sorry for that - my brain usually melts at this time of night! Corrected in my post.2011-04-09