The non-rotated ellipse centered at the origin has equation $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
We can show via implicit differentiation that the equation of the tangent at $\left (x_0,y_0 \right )$ is $$\frac{xx_0}{a^2}+\frac{yy_0}{b^2}=1$$
In other words, we have simply 'borrowed' an $x$ and turned it into $x_0$ and similarly for $y_0$.
Is there a deeper reason for this being the case, or is this just a mere coincidence?
On
See implicit differentiation.
Find the equation of the tangent line to the ellipse at the given point
It happens that $
dy / dx = (-b^2/a^2) \cdot x/y$
The rest is just finding the slope at the point $(x_0, y_0)$
So I guess you can call it a coincidence, yes.
In my opinion, there is a deeper reason.
For plainly understanding it, one has to work in the associated projective space and do projective geometry (I compare the introduction of projective geometry in classical geometry as the introduction of complex numbers when you have worked only with real numbers: it largely widens the scope).
In fact, far from a course on projective geometry, I will only scrap at its surface, by taking an example and show that the association of a point on a conic curve with its tangent line can be enlarged.
Consider the following second degree polynomial in two variables,
$$2x^2+2y^2+2xy-6y=0 \ \ (1)$$
First step: set $x=X/T, y=Y/T \ \ (*)$. Why that ? Because this (mostly dumb) variable $T$ will allow (by setting it to zero) to speak about and work on elements (points in particular) said "at infinity".
When expressions (*) are plugged into (1), and after a multiplication by $T^2$, one gets:
$$2X^2+2Y^2+2XY-6YT=0 \ \ (2)$$
One says that (**) is the homogeneized version of (1) (all terms are now "second degree terms"). What is the change ? Only a supplementary $T$ (another example: if we had worked on $x^2 -2y -8$, the result would have been $X^2 - 2YT - 8T^2$.)
One says also that (2) is a quadratic form, with an associated bilinear form obtained by applying the following 2 rules:
If you meet a square like $X^2$, $Y^2$, $T^2$, replace it resp. by $XX', YY', TT'$;
If you meet a "mixed" term, like $2XY, 2XT$ or $2YT$ replace it resp. by $XY'+YX', XT'+X'T, YT'+Y'T$. (mnemonics: if you cancel the "prime(s)" you come back to the initial expressions).
For example, the bilinear form associated with the quadratic form (2) is
$$2XX'+2YY'+ XY' + YX'- 3(YT'+Y'T)=0 \ \ (3)$$
Often, one make the $T$'s disappear (in fact by setting $T=T'=1$), and switching back to lower case letters like this (we have also replaced $X', Y'$ by $x_0,y_0$)
$$2xx_0+2yy_0+ xy_0 + yx_0 - 3(y+y_0)=0 \ \ (4)$$
Now, we are ready for the application we had in view:
either point $P_0(x_0,y_0)$ is on the conical curve (C) (see figure) and (4) is the equation of the tangent in this point to the conic curve,
or $P_0(x_0,y_0)$ is not on (C). In this case, $P_0$ will be called a "pole" and (4), which still is the equation of a straight line, will be called the "polar line" associated with this pole. This is the natural generalization of the concept of tangent. One speaks of "the duality between points and straight lines" (it is one of the main historical origins of the concept of duality), being given a "reference" conical curve.
Fortunately, when the pole $P_0$ is outside the reference conical curve (C), there is a nice way to obtain geometrically its polar line: draw the two tangent lines to $(C)$. Let $Q_0$ and $Q_1$ be their contact points with (C); then the polar line is the straight line $Q_0Q_1$ (see examples on the figure).
Remark; there is a very useful presentation of these issues using matrices, for example (3) can be written in this way:
$$U'^TAU \ = \ \begin{pmatrix} X' & Y' & T' \end{pmatrix} \begin{pmatrix} 2 & 1 & 0\\ 1 & 2 & -3\\ 0 & -3 & 0 \end{pmatrix} \begin{pmatrix} X\\ Y\\ T \end{pmatrix}$$
where $A$ is a symmetric matrix.
Edit: In this spirit, the duality pole $\rightarrow$ polar is the following transformation $AU=U'$:
$$\begin{pmatrix} x_0\\ y_0\\ 1 \end{pmatrix} \rightarrow \begin{pmatrix} 2 & 1 & 0\\ 1 & 2 & -3\\ 0 & -3 & 0 \end{pmatrix} \begin{pmatrix} x_0\\ y_0\\ 1 \end{pmatrix} $$
which gives coefficients $u,v,w$ of an equation $ux+vy+w=0$ of the polar line of $(x_0,y_0)$.
In a reverse sense, if you have the coefficients $U=(u,v,w)^T$ of a straight line, it will be the polar of the pole defined by $A^{-1}U$ ; some caution is necessary ; in fact, as coefficients $u,v,w$ are known up to a multiplicative factor, one has to find the right factor in order that the value of the 3rd coefficient is one (in the ordinary cases ; if this operation is impossible, it means that this straight line passes through the center of (C))