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I am quite new to differential equations and derivatives. I want to derive an differential form for equation of an ellipse. If i start with an ordinary ellipse equation

\begin{equation} \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \end{equation}

How do i derive it then to get this form

$ -\frac{dx}{dy} = \frac{a^2}{b^2} \frac{y}{x} $

I would need an equation and some brief explanation on the procedure.

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    Have you encountered "implicit differentiation?" Here is a helpful link: http://www.sosmath.com/calculus/diff/der05/der05.html2012-04-05

2 Answers 2

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The equation $ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \tag{1}$ has two variables: $\{ x, y \}.$ By "derive," it seems that you mean $ \frac{dx}{dy}.$

Well, differentiating equation $(1)$ w.r.t $y,$ we get: $ \frac{d}{dy} \frac{x^2}{a^2} + \frac{d}{dy} \frac{y^2}{b^2} = \frac{d}{dy} 1 \tag{2}$

First, note that $\dfrac{d}{dy} 1 = 0,$ $\dfrac{d}{dy} y = 1,$ and $\dfrac{d}{dy} f^2 = 2 f \dfrac{df}{dy}.$

So

  1. $ \dfrac{d}{dy} \dfrac{x^2}{a^2} = 2 \dfrac{x^{2-1}}{a^2} \dfrac{dx}{dy}$

  2. $ \dfrac{d}{dy} \dfrac{y^2}{b^2} = 2 \dfrac{y^{2-1}}{b^2} \dfrac{dy}{dy} $

In other words, equation $(2)$ becomes:

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

The rest is simple algebra, you can isolate $\dfrac{dx}{dy}$ one side, and get: $ -\frac{dx}{dy} = \frac{a^2}{b^2} \frac{y}{x} $

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    Thank you, this is a good e$x$planation.2012-04-10
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The answer you want is actually not the differential equation of the family of ellipse. A differential equation is free of arbitrary constants like $a$ and $b$. Since there are two arbitrary constants, you need to differentiate 2 times (the order of the differential equation should be 2). The answer you want is just the negative of slope of normal at any point $(x,y)$ on the ellipse.