**Original Question:**

Let , where , be a point on the ellipse . Show that the tangent line to the ellipse passing through is .

**Suggested Solution:**

This is not a difficult question to tackle considering the fact that is on the circle. Some basic set-ups: and . Performing implicit differentiation and some further algebraic manipulations, we obtain the gradient of the tangent at : . Substituting the gradient of the tangent into the tangent equation (where m refers to the gradient) to obtain the equation, simplify and we can get . After that, we can easily get the answer by realising .

**A rather routine question so far, but what if the point is not on the circle, and it is placed outside, or inside the circle instead? What is the geometrical meaning of the equation ? **

**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 to the circle. At the point on the circle, the tangent of the circle L2 is . Then, since is on L2, we have . Based on the equation, we can conclude that is on the line .

Similarly, at the point on the circle, the tangent of the circle L1 is .. Then, since is on L1, we have . Based on the equation, we can conclude that is on the line

With and on L, we can conclude that the geometrical meaning of the equation is the line passing through the two points on the circle such that the line between and the point is tangential to the circle.

**Finally, the inside scenario, the most tricky case**

Locate two points and on the circle such that BC (L3) passes through A, and A is the mid-point of BC, i.e. . Let the tangents (L1 and L2) passing through B and C intersect outside the circle at ** **D(m,n).

By the previous discussion when is outside the circle, the equation of L3 can be written as . Since A is on L3, then we have . Hence D is on the line L: . Re-arranging the equation, we get

Hence, the gradient of L is

Consider the line L3, since it passes through and , we have (*) and (**). (*)-(**): , which is equivalent to

From the above equation, we can simply the gradient of L3: = gradient of L

That is to say, the geometrical meaning of 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.