Consider the ellipse with canonical equation referring to the axes and origin of coordinates in the center of the ellipse, i.e., with equation
$$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\qquad \text{$(a,\,b>0)$ semiaxes.}$$
With partial derivatives, if we have in the plane a curve $C$ of implicit equation $f(x,y)=0$, where $f$ is continuous with prime partial derivatives $\partial_x f$ and $\partial_y f$ also continuous, given a point $(x_0, y_0)\in C$ on the noncritical curve for $f$ (i.e., such that $\nabla f(x_0,y_0)\ne(0,0)$), it is known that the equation of the tangent line to the curve at the point $(x_0,y_0)$ in question is given by
$$\partial f_x(x_0,y_0)(x-x_0)+\partial f_y(x_0,y_0)(y-y_0)=0.$$
In our case, therefore, we have that the equation of the tangent is
$$\dfrac{2x_0}{a^2}\,(x-x_0)+\dfrac{2y_0}{b^2}\,(y-y_0)=0,$$
from which with obvious steps
$$\dfrac{x_0\,x}{a^2}+\dfrac{y_0\,y}{b^2}-\dfrac{x_0^2}{a^2}-\dfrac{y_0^2}{b^2}=0,$$
That is, since $(x_0,y_0)$ lies on the ellipse
$$\boxed{\dfrac{x_0\,x}{a^2}+\dfrac{y_0\,y}{b^2}=1.} \tag 1$$
In this website there is also a long tedious proof where the Italian language is just barely perceptible.
Is there a proof of the $(1)$, using homoteties (dilations) for example, that is simpler and more immediate? Partial derivatives are not studied in high school.
Best Answer
With the transformation $x=aX, y=bY$, the equation of the ellipse becomes the circle $X^2+Y^2=1$.
The tangent to the circle in one of its points can be found by geometric methods (tangent line is orthogonal to the radius) and is given by $X_0X+Y_0Y=1$.
By transforming back to the original coordinates we obtain the desired result.
The slope of the radius is $$ m_r = \frac{Y_0-0}{X_0-0} = \frac{Y_0}{X_0} $$ then the slope of the tangent line is $m_t = -1/m_r = -X_0/Y_0$, so that the tangent line has equation \begin{align} & Y-Y_0 = m_t(X-X_0) \\ & Y-Y_0 = -\frac{X_0}{Y_0}(X-X_0) \\ & (X-X_0)X_0+(Y-Y_0)Y_0 = 0 \\ & X_0X+Y_0Y = X_0^2+Y_0^2 \\ & X_0X+Y_0Y = 1 \\ \end{align}
Finally, given that $(X,Y)=(x/a,y/b)$ and $(X_0,Y_0)=(x_0/a,y_0/b)$ we have $$ \frac{x_0x}{a^2}+\frac{y_0y}{b^2} = 1 $$