Derivative of an ellipse

Question

I'm in my last year of high school and I'm currently studying on conics. Usually to get the centre of an ellipse for example I use the canonical form to get the following form $((x+k)/a)^2 + ((y+k)/b)^2=1$ then consider the $x+k$ and the $y+k$ at the coordinates of the centre. My Math teacher showed us a method to find this center without using the canonical form, just by using derivatives on the initial form which is: $ax^2 +by^2 + cx + dy + f = 0$. So his method consists of deriving this equation once with respect to $x$ and consider the $y$ a constant then have an equation which is $2ax + c = 0$ and the derive the same equation with respect to $y$ and considering $x$ a constant and having: $2by + d = 0$, and finally by solving these two equations he will have the center of the ellipse. Can I have an explanation on why he used this method? like what do the derivatives have to do with the ellipse, where did this method come from.

Sorry for any English mistakes and for the wordiness, Thank you in advance.

Answer 1

The key idea is that the derivative of a function is zero when it is at its maximum or minimum - for example, the parabola $x^2$ has derivative zero exactly when $x = 0$, which is the bottom point of the parabola. The reason for this is that when a function reaches its maximum or minimum, it must (instantaneously) level out, not going up or down, which means it has zero derivative at that point.

Taking those derivatives you describe, then, acquires the $x$ and $y$ values of the points on the ellipse that are "flat"; those are the horizontal and vertical "end points" of the ellipse, which are the points along the ellipse's axes.

Answer 2

The derivative with respect to $x$ while holding $y$ constant will give you the equation whose solution is the $x$-coordinate of the maximum/minimum $y$-value of the ellipse. Similarly, the derivative with respect to $y$ while holding $x$ constant will give you the equation whose solution is the $y$-coordinate of the maximum/minimum $x$-value of the ellipse. Since the ellipse has some symmetry about its axes, the coordinates of the $x$ and $y$ minima/maxima will correspond to the center point at $(x,y)$.

Answer 3

Consider the left-hand side of the ellipse formula as a two-parameter function, $$f(x,y) = ax^2 +by^2 + cx + dy + f.$$ You can then use this function to plot a surface in three-dimensional Cartesian coordinates using the formula $z = f(x,y)$.

Assuming you started with the usual form of an ellipse equation (with $a > 0$ and $b > 0$), the surface described by $z = f(x,y)$ is a "bowl-shaped" surface. Assuming the original equation $f(x,y) = 0$ described an ellipse, the "bowl" intersects the $x,y$ plane, and the "bottom" of the bowl is directly beneath the center of the ellipse.

Now by choosing any constant value of $y$, we take a vertical cross-section of the "bowl" parallel to the $x,z$ plane. This cross-section is a parabola. Since we oriented the ellipse with its axes parallel to the $x$ and $y$ axes (by not having a non-zero $xy$ term), and because the bowl has reflective symmetry across the vertical planes through the axes of the ellipse, the bottom of the parabolic cross-section is on the axis of the ellipse parallel to the $y$-axis. All $x$ coordinates on this axis of the ellipse, including the $x$ coordinate of the center of the ellipse, are the same, so by taking any arbitrary cross-section and finding the bottom of it (by setting the derivative of the parabolic function to zero) we find the $x$ coordinate of the center of the ellipse. The cross-section does not even have to intersect the ellipse in order for this to work, because the plane of symmetry of the bowl follows the axis of the ellipse as far as you like in either direction.

A similar trick holding the $x$ value constant, taking a cross-section parallel to the $y,z$ plane, finds the $y$ coordinate of the center.

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In fact, a small extension of this method works even for the more general ellipse equation, $ax^2 +by^2 + gxy + cx + dy + f = 0,$ which can describe an ellipse in any orientation, not just one with its axes parallel to the $x$ and $y$ axes. The procedure is just slightly more complicated, because now the "bottom" of each parabolic cross-section for a constant $y$ value depends on which value of $y$ is chosen; but the "bottom" of every such cross-section lies on a line through the two points on the ellipse at minimum and maximum $y$ values. The derivative when we hold $y$ constant is the equation of that line. Likewise, holding $x$ constant and taking the derivative with respect to $y$, we get the equation of a line through the points on the ellipse at minimum and maximum $x$ values. By symmetry, these lines cross at the center of the ellipse. So to find the center, we take the equations of the two lines as a system of simultaneous linear equations, and solve it.