# The Geometrical Meaning of the Tangent of an Eclipse

Original Question:
Let $(x_{0},y_{0})$, where $y_{0} \neq 0$, be a point on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. Show that the tangent line to the ellipse passing through $(x_{0},y_{0})$ is  $\frac{x_{0}x}{a^{2}}+\frac{y_{0}y}{b^2}=1$.

Suggested Solution:
This is not a difficult question to tackle considering the fact that $(x_{0},y_{0})$ is on the circle. Some basic set-ups:  $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$  and $b^2x^2+a^2y^2=a^2b^2$ . Performing implicit differentiation and some further algebraic manipulations, we obtain the gradient of the tangent at $(x_{0},y_{0})$:  $\frac{dy}{dx}=\frac{-b^2x}{a^2y}$ . Substituting the gradient of the tangent into the tangent equation $y-y_{0}=m\times(x-x_{1})$ (where m refers to the gradient) to obtain the equation, simplify and we can get $\frac{x_{0}x}{a^2}+\frac{y_{0}y}{b^2}=\frac{x_{0}^2}{a^2}+\frac{y_{0}^2}{b^2}$. After that, we can easily get the answer $\frac{x_{0}x}{a^{2}}+\frac{y_{0}y}{b^2}=1$ by realising $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$.

A rather routine question so far, but what if the point $(x_{0},y_{0})$ is not on the circle, and it is placed outside, or inside the circle instead? What is the geometrical meaning of the equation  $\frac{x_{0}x}{a^{2}}+\frac{y_{0}y}{b^2}=1$

Let us consider the outside scenario first. (In the diagrams below, I use circles to represent eclipses. Sorry for the inaccuracy.)

According to the diagram on the right, draw two tangents (L1 and L2) from the point $(x_{0},y_{0})$ to the circle.  At the point $B (x_{1},y_{1})$ on the circle, the tangent of the circle L2 is $\frac{x_{1}x}{a^{2}}+\frac{y_{1}y}{b^2}=1$. Then, since $(x_{0},y_{0})$ is on L2, we have $\frac{x_{1}x_{0}}{a^{2}}+\frac{y_{1}y_{0}}{b^2}=1$. Based on the equation, we can conclude that $B (x_{1},y_{1})$ is on the line $\frac{x_{0}x}{a^{2}}+\frac{y_{0}y}{b^2}=1$.

Similarly, at the point $C (x_{2},y_{2})$ on the circle, the tangent of the circle L1 is $\frac{x_{2}x}{a^{2}}+\frac{y_{2}y}{b^2}=1$.. Then, since $(x_{0},y_{0})$ is on L1, we have $\frac{x_{2}x_{0}}{a^{2}}+\frac{y_{2}y_{0}}{b^2}=1$. Based on the equation, we can conclude that $B (x_{2},y_{2})$ is on the line $\frac{x_{0}x}{a^{2}}+\frac{y_{0}y}{b^2}=1$

With $B (x_{1},y_{1})$ and $C (x_{2},y_{2})$ on L, we can conclude that the geometrical meaning of the equation $\frac{x_{0}x}{a^{2}}+\frac{y_{0}y}{b^2}=1$ is the line passing through the two points on the circle such that the line between $(x_{0},y_{0})$ and the point is tangential to the circle.

Finally, the inside scenario, the most tricky case

Locate two points $B (x_{1},y_{1})$ and $C (x_{2},y_{2})$ on the circle such that BC (L3) passes through A$(x_{0},y_{0})$, and A is the mid-point of BC, i.e. $x_{0}=\frac{1}{2}(x_{1}+x_{2}); y_{0}=\frac{1}{2}(y_{1}+y_{2})$ . Let the tangents (L1 and L2) passing through B and C intersect outside the circle at  D(m,n).

By the previous discussion when $(x_{0},y_{0})$ is outside the circle, the equation of L3 can be written as $\frac{mx}{a^{2}}+\frac{ny}{b^2}=1$. Since A is on L3, then we have $\frac{mx_{0}}{a^{2}}+\frac{ny_{0}}{b^2}=1$.  Hence D is on the line L: $\frac{x_{0}x}{a^{2}}+\frac{y_{0}y}{b^2}=1$. Re-arranging the equation, we get $y=-\frac{b^2x_{0}}{a^2y_{0}}x+\frac{b^2}{y_{0}}$

Hence, the gradient of L is $-\frac{b^2x_{0}}{a^2y_{0}}$

Consider the line L3, since it passes through $B (x_{1},y_{1})$ and $C (x_{2},y_{2})$, we have $\frac{x_{1}}{a^{2}}+\frac{y_{1}}{b^2}=1$ (*) and $\frac{x_{2}}{a^{2}}+\frac{y_{2}}{b^2}=1$ (**). (*)-(**): $\frac{x_{1}^2-x_{2}^2}{a^2}+\frac{y_{1}^2-y_{2}^2}{b^2}=0$, which is equivalent to $\frac{(x_{1}+x_{2})(x_{1}-x_{2})}{a^2}+\frac{(y_{1}+y_{2})(y_{1}-y_{2})}{b^2}=0$

From the above equation, we can simply the gradient of L3: $\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\frac{-b^2(x_{1}+x_{2})}{a^2(y_{1}+y_{2})}=\frac{-b^2(2x_{0})}{a^2(2y_{0})}=-\frac{b^2x_{0}}{a^2y_{0}}$ = gradient of L

That is to say, the geometrical meaning of $\frac{x_{0}x}{a^{2}}+\frac{y_{0}y}{b^2}=1$ is a line parallel to BC and passes through the intersection of the tangents passing through B and C.

Note that my solution for the final case may not be perfect. What if I cannot construct ABC according to the manner I expect?

More specially, there could be something more to show: (1)Draw any secant BC of the ellipse passing through A (A is not necessarily the midpoint of BC), and draw the tangent lines to the ellipse at B and C.  Then these two lines intersect at a point on the line L.  (2)Conversely, pick any point D on L, and draw the tangent lines to the ellipses.  Let B and C be the tangent points on the ellipse.  Then D lies on BC. Based on the two conditions we can reach the same  conclusion. I believe that this method should be more credible, but considerably more tedious as well.

Feel free to leave your comments if you have critics or alternatives.

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