Originally posted by George W. Bush
CM: Is there a easy way to solve the following question without extensive algebra?
The points P and Q lie on an ellipse (standard eqn). The tangents and P and Q intersect at point T. The midpoint of PQ is the point M. Prove that OM extended (where O is origin) passes through the point T.
I've been giving this one some thought - most obvious approaches will result in an algebra bash of unnecessary complexity. I don't see a straight geometric approach, however here is a reasonable algebraic approach.
Let's assume that the point T is (x<sub>0</sub>, y<sub>0</sub>), in which case m<sub>OT</sub> = y<sub>0</sub> / x<sub>0</sub>. Since any two parallel lines with a common point are the same line, it will be sufficient to prove that m<sub>OM</sub> = y<sub>0</sub> / x<sub>0</sub>.
Now, the chord of contact from T is xx<sub>0</sub> / a<sup>2</sup> + yy<sub>0</sub> / b<sup>2</sup> = 1 _____(1)
It meets the ellipse at the points P and Q, and by making y the subject of (1), and putting it into the equation of the ellipse, we get a quadratic equation for the x coordinates of P and Q:
x<sup>2</sup> / a<sup>2</sup> + (1 / b<sup>2</sup>y<sub>0</sub><sup>2</sup>) * (b<sup>2</sup> - b<sup>2</sup>xx<sub>0</sub> / a<sup>2</sup>)<sup>2</sup> = 1
ie x<sup>2</sup>[(1 / a<sup>2</sup>) + (b<sup>2</sup>x<sub>0</sub><sup>2</sup> / a<sup>4</sup>y<sub>0</sub><sup>2</sup>] - 2b<sup>2</sup>xx<sub>0</sub> / a<sup>2</sup>y<sub>0</sub><sup>2</sup> - 1 + b<sup>2</sup> / y<sub>0</sub><sup>2</sup> = 0
We don't care what the solutions are, but we do care what their average is, as it is the x coordinate of M. Thus,
x coordinate of M = -B / 2A = (2b<sup>2</sup>x<sub>0</sub> / a<sup>2</sup>y<sub>0</sub><sup>2</sup>) / 2[(1 / a<sup>2</sup>) + (b<sup>2</sup>x<sub>0</sub><sup>2</sup> / a<sup>4</sup>y<sub>0</sub><sup>2</sup>] = a<sup>2</sup>b<sup>2</sup>x<sub>0</sub> / (a<sup>2</sup>y<sub>0</sub><sup>2</sup> + b<sup>2</sup>x<sub>0</sub><sup>2</sup>)
By making x the subject of (1), we can similarly find an equation whose solution is the y coordinate of P and Q, and use it to show that the y coordinate of M is a<sup>2</sup>b<sup>2</sup>y<sub>0</sub> / (b<sup>2</sup>x<sub>0</sub><sup>2</sup> + a<sup>2</sup>y<sub>0</sub><sup>2</sup>)
Thus, m<sub>OM</sub> = [a<sup>2</sup>b<sup>2</sup>y<sub>0</sub> / (b<sup>2</sup>x<sub>0</sub><sup>2</sup> + a<sup>2</sup>y<sub>0</sub><sup>2</sup>)] / [a<sup>2</sup>b<sup>2</sup>x<sub>0</sub> / (a<sup>2</sup>y<sub>0</sub><sup>2</sup> + b<sup>2</sup>x<sub>0</sub><sup>2</sup>)] = y<sub>0</sub> / x<sub>0</sub>, as required.
Note that this approach requires both x<sub>0</sub> and y<sub>0</sub> to both be non-zero. However, if either of these is zero, then the point T lies on an axes, and symmetry dictates that M must lie on the same axis - try drawing the diagram if you're not convinced - and thus the theorem is obviously true for these cases as well.
