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?
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 $$(x,y)=\frac1{\sqrt{a^2m^2+b^2}}(a^2m,\ b^2),\ \mathbb R\ni m>0$$