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Suppose I have have an ellipse $ \frac{x^2}{a^2 } + \frac{y^2}{b^2} = 1$, let $\frac{dy}{dx}=m$, I want to find the expression for write the points on the first quarter of ellipse as $\left(x(m),y(m) \right)$; I want to parameterize the coordinates using the slope of the curve.

How would I go about doing this?

I thought to take the derivative of ellipse equation:

$$ \frac{x}{a^2} \frac{dx}{dm} + \frac{y }{b^2} \frac{dy}{dm}= 0$$

Is what I find, but how do I move from the above into explicit parameterizations of $(x,y)$ in terms of the slope?

P.s: I know an easy way to do this is to put $(x,y)=(a \cos t , b \sin t)$ then $m= \frac{b}{a} \tan t$ and life is easy but this was easy because of our choice of parameterization.

Could we move directly from the $(x,y)$ equation into a parameterization in terms of slope for the chosen coordinates?

tryst with freedom
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  • Just to make sure I'm on the same page, when you say "first quarter of the ellipse", are you referring to the points $(x,y)=(a\cos t,b\sin t)$ for $t\in[0,\pi/2]$? – Alann Rosas Aug 17 '21 at 05:34
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    just the (x,y) pairs which solve the ellipse equation with 0<x,y<1 , the (acost , b sin t ) is a choice of parametrization @AlannRosas – tryst with freedom Aug 17 '21 at 05:36

1 Answers1

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Here's a non-geometrical approach. Differentiating the equation of the ellipse and solving for $y$ using $y'(x)=m$ gives :

$$y=-\left(\frac{b^2}{a^2}\right)\bigg(\frac{x}{m}\bigg)$$

Putting this back into the equation of the ellipse and solving for $x$ (with the positive square root)gives :

$$x=\frac{a^2m}{\sqrt{a^2m^2+b^2}}$$

Using this to solve for $y$ gives :

$$y=\frac{b^2}{\sqrt{b^2+a^2m^2}}$$

Thus you can use the parametrization $$(x,y)=\frac1{\sqrt{a^2m^2+b^2}}(a^2m,\ b^2),\ \mathbb R\ni m>0$$

Yuzuriha Inori
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  • Note that this can be extended to all $m \in \mathbb{R}$; negative values of $m$ correspond to points on the ellipse in Quadrant II. – Michael Seifert Aug 17 '21 at 14:08