# How to derive a differential equation of an ellipse

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

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$

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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An ellipse does not hace a differential equation... What exactly are you trying to do? – Mariano Suárez-Alvarez Apr 5 '12 at 22:56
I think perhaps "derive" means take derivatives? – Alex R. Apr 5 '12 at 23:01
Have you encountered "implicit differentiation?" Here is a helpful link: sosmath.com/calculus/diff/der05/der05.html – Alex R. Apr 5 '12 at 23:03

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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Oh. I forget to add, $a, b$ are constants, and the differential operator is linear. – user2468 Apr 5 '12 at 23:40
Thank you, this is a good explanation. – 71GA Apr 10 '12 at 7:17

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.

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