Formula for the tangent to an ellipse at a given point (splitting formula)

Question

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.

Answer 1

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.

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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 $$ \frac{x_0x}{a^2}+\frac{y_0y}{b^2} = 1 $$

Answer 2

Let $$ax^2+2bxy+cy^2+2dx+2ey+f=0$$ be a nondegenerate conic, with a point $(x_0,y_0)$ on it.

Translate the equation to the origin

$$0=a(x+x_0)^2+2b(x+x_0)(y+y_0)+c(y+y_0)^2+2e(y+y_0)+2d(x+x_0)+f\\=[ax^2+2bxy+cy^2]+\\2(axx_0+b(xy_0+x_0y)+cyy_0+dx+ey)\\+[ax_0^2+2bx_0y_0+cy_0^2+2dx_0+2ey_0+f]$$

where the first pair of square brackets don't contribute to the tangent since we're at the origin and these terms are of degree two. And the second pair of square brackets are zero since $(x_0,y_0)$ is on the conic. Then we translate (half of it) back $$a(x-x_0)x_0+b((x-x_0)y_0+x_0(y-y_0))+c(y-y_0)y_0+d(x-x_0)+e(y-y_0)\\=[-ax_0^2-2bx_0y_0-cy_0^2-2dx_0-2ey_0-f]+\\axx_0+b(xy_0+x_0y)+cyy_0+d(x+x_0)+e(y+y_0)+f=0.$$

This method is used in algebraic geometry, near singular points; search for "the tangent cone". But it also works to find the tangent at regular points, and I don't see why a high school student couldn't get it, if they're looking at degree two equations anyway and with the above as the definition of a tangent: if you're at the origin, throw away higher order terms keeping the linear terms, the constant term already having vanished as the origin is on the curve.