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I have a relation $$\frac{(x\cos(\tan^{-1}(x_1))-y\sin(\tan^{-1}(x_1)))^2}{a^2}+\frac{(y\cos(\tan^{-1}(x_1))+x\sin(\tan^{-1}(x_1)))^2}{b^2}=1$$

$a^2$ and $b^2$ are both functions solely in terms of $x_1$. For various real number values of $x_1$, the function on the Cartesian coordinate plane is that of a rotated ellipse. I want to find what the equation of the derivative of that ellipse trends towards as $x_1$ approaches toward infinity. I believe to get that I need the limit as $x_1$ approaches infinity of some sort of function in both $x_1$ and $x$ in which I can substitute $x_1$ and get the equation of the derivative of the ellipse that is formed by that value of $x_1$. I presume to get that I need to take some type of derivative, however, I do not know what type that might be, or whether it is even possible to get the derivatives of the ellipses simply by substituting a variable.

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  • $\begingroup$ The parentheses are unbalanced in the second term. I suspect that the term should be $\frac{(y\cos(\tan^{-1}(x_1))+x\sin(\tan^{-1}(x_1)))^2}{b^2}$, but I can't be certain. $\endgroup$
    – Alex Jones
    Dec 4, 2018 at 18:56
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    $\begingroup$ That’s right, thank you for changing that! $\endgroup$
    – H Huang
    Dec 4, 2018 at 19:02
  • $\begingroup$ What do you mean by "the derivative of an ellipse?" Do you mean the derivative of $y$ as an implicit function of $x$? What is the derivative of $x^2+y^2=1?$ $\endgroup$
    – saulspatz
    Dec 4, 2018 at 19:19

1 Answer 1

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First, it's important to clear up some terminology, since it really informs the approach here. The equation

$$\frac{[x\cos(\tan^{-1}s)-y\sin(\tan^{-1}s)]^2}{a(s)^2}+\frac{[y\cos(\tan^{-1}s)+x\sin(\tan^{-1}s)]^2}{b(s)^2}=1, \text{ for } s\in\mathbb{R}$$

is not a function. If you attempt to solve for either of the variables to write it as a function, you will find that you will be unable to fully describe the all the points as the graph of a function of either $x$ or $y$. Graphically, this is equivalent to the fact that there is at least 1 vertical and 1 horizontal line which intersects any ellipse.

The ellipse is a closed, non-self-intersecting, planar curve, and in general there are 2 ways to get the derivative. The first is by calculating the derivative directly via implicit differentiation, and the second is by parameterizing the curve as a function of 1 variable and taking the derivative of that function. These two derivatives $y'(x,y)$ and $F'(\theta) = (F'(\theta)_x,F'(\theta)_y)$ will be different, but are related by $y'(F(\theta)) = \frac{F'(\theta)_y}{F'(\theta)_x}$, where $F(\theta) = (x,y)$ is the parameterization.


Let's try implicit differentiation. Before going ahead, let's make the substitution $c(s) = \cos(\tan^{-1}s)$ and $d(s) = \sin(\tan^{-1}s)$, and for simplicity we may drop the explicit $s$ dependence when referencing $a,b,c,d$. With this, the equation becomes

$$ \frac{(cx-dy)^2}{a^2}+\frac{(dx+cy)^2}{b^2}=1 $$

We take the derivative of both sides with respect to $x$, considering $y$ as a (possibly multi-valued) function of $x$. Thus, $\frac{\text d}{\text dx} f(y) = f'(y)y'$ for any differentiable function $f$. This gives us the equation

$$ \frac{(cx-dy)(c-dy')}{a^2} + \frac{(dx+cy)(d+cy')}{b^2} = 0 $$

which can be solved for $y'$ to give

$$ y'(x,y) = \frac{b^2c(cx-dy)+a^2d(dx+cy)}{b^2d(cx-dy)-a^2c(dx+cy)} $$

As $s\to\pm\infty$, $c\to 0$ and $d\to \pm 1$. Thus, we can write the limit

$$ \lim\limits_{s\to\pm\infty} y'(x,y;s) = -\frac{a(\pm\infty)^2}{b(\pm\infty)^2}\frac{x}{y}$$

This gives an expression for the derivative in terms of points $(x,y)$ on the ellipse. You need to be careful with this, though, because it will also give you a value for points off the ellipse, which are not meaningful. If you want an equation which only relates points on the ellipse to their derivative, you can solve the original ellipse equation for $y$, splitting into two functions for the upper and lower halves, and substituting each of those for $y$ to get the two corresponding expressions for the derivatives. I'll leave this to you if you want to do it.


Alternatively, you can write the curve as the graph of a function of 1 variable. Before doing this, we will make the substitution $\tan^{-1}s = \phi$, and so define the functions $A(\phi) = a(\tan\phi)$ and $B(\phi) = b(\tan\phi)$, and for simplicity we may drop the explicit $\phi$ dependence when referencing $A,B$. Then, define the function

$$F(\theta) = (A\cos\theta\cos\phi + B\sin\theta\sin\phi, - A\cos\theta\sin\phi + B\sin\theta\cos\phi)$$

We can take the derivative easily

$$F'(\theta) = (-A\sin\theta\cos\phi + B\cos\theta\sin\phi, A\sin\theta\sin\phi + B\cos\theta\cos\phi)$$

Then we can write the derivative as a function of $\theta$ by dividing

$$y'(F(\theta)) = \frac{B\cos\theta\cos\phi+A\sin\theta\sin\phi }{ B\cos\theta\sin\phi-A\sin\theta\cos\phi }$$

As $s\to\pm\infty$, $\phi\to\pm\frac{\pi}{2}$, so we can write the limit

$$\lim\limits_{\phi\to\pm\frac{\pi}{2}} y'(F(\theta);\phi) = \frac{A(\pm\frac{\pi}{2})}{ B(\pm\frac{\pi}{2})}\tan\theta $$


Equivalence of these two expressions can be seen as follows. When $\phi\to\pm\frac{\pi}{2}$, $F(\theta)\to(\pm B(\pm\frac{\pi}{2})\sin(\theta),\mp A(\pm\frac{\pi}{2})\cos(\theta))$. Substituting this into $y'(x,y)$, noting that $a(\pm\infty) = A(\pm\frac{\pi}{2})$ and $b(\pm\infty) = B(\pm\frac{\pi}{2})$, gives the expression

$$y'(F(\theta)) = -\frac{A(\pm\frac{\pi}{2})^2}{B(\pm\frac{\pi}{2})^2}\frac{\pm B(\pm\frac{\pi}{2})\sin(\theta)}{\mp A(\pm\frac{\pi}{2})\cos(\theta)} = \frac{A(\pm\frac{\pi}{2})}{ B(\pm\frac{\pi}{2})}\tan\theta$$

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