A parabola is drawn such that each vertex of a given triangle is the pole of the opposite side;show that the focus of the parabola lies on the nine point circle of the triangle and that the orthocentre of the triangle formed by joining the middle points of the sides lies on the directix.
This question from sl loney (parabola) page no 191 exercise 28 question 30.
I tried to start with points (s,t) (u,v) (x,y) and tried to take their poles..but it became very ugly.
I'm pretty sure that you can find the answer in the following book.
"Text Book Of Circles And Parabola" A.K.Sharma.
Solution
The circle circumscribing the triangle formed by joining the middle points of the sides of the given triangle will be the nine point circle. As the circle circumscribing the triangle formed by the three tangents on the parabola always passes through the focus of the parabola (Ref. Art.179), hence, we have to prove that the lines joining the mid-points of the given triangle are the tangents on the parabola.
Let the equation of the parabola be $y^2=4ax. (1)$
Suppose the given triangle is $ABC$ and let $PQ$ be the chord of contact of A w.r.t. (1). Again, as $BC$ is given to the polar of $A$ w.r.t. (1), BC and PQ must lie on the same line. Hence the line joining the mid. points of AP and AQ also passes through the mid.points of AB & AC.
Suppose the co-odinates of P and Q be $(a{t_1}^2,2at_1)$ and $(a{t_2}^2, 2at_2)$ respectively so that the co-ordinates of the point of intersection of the tangents at P and Q i.e. A are $[at_1t_2,a(t_1+t_2)]$
Hence, Co-ordinates of mid-point of AP will be $$\left\{\frac{at_1(t_1+t_2)}{2},\frac{a(3t_1+t_2)}{2}\right\}$$ Similarly, the co-ordinates of mid-point of AQ will be $$\left\{\frac{at_2(t_1+t_2)}{2},\frac{a(t_1+3t_2)}{2}\right\}$$ The line joining the mid-points of AP and AQ is $$y-\frac{a(3t_1+t_2)}{2}=\frac{\frac{a(3t_1+t_2)}{2}-a\frac{(t_1+3t_2)}{2}}{\frac{at_1(t_1+t_2)}{2}-\frac{at_2(t_1+t_2)}{2}}\left\{x-\frac{at_1(t_1+t_2)}{2}\right\}$$ $$\Rightarrow y-\frac{2}{t_1+t_2}x+\frac{a(t_1+t_2)}{2}$$ This is a tangent on the parabola as it is of the form $y=mx+a/m$ Hence proved.
Again the ortho-centre of the triangle fromed by the tangents is on directrix (art. 181) and we have proved that the lines joining the mid-points of the sides of the triangle are tangents on the parabola.
Hence the ortho-centre of the triangle formed by joining the mid. Point of the sides of the triangle lies on directrix of the parabola. Proved.