From a point $(h,k)$, 3 distinct normals can be drawn to the parabola $y^2=4ax$ and the feet of these normals are $P(t_1),Q(t_2),R(t_3)$. Find the centroid of $\Delta PAQ$
My method of solving this is extremely rudimentary, but it got me very close to the answer.
I assumed the given point to like on the X axis for the sake of simplicity. This meant that one of the feet was the origin
The equation of normal was
$$k=-ht+2at+at^3$$
Since k =0
$$-h+2a+at^2=0$$
$$at^2=h-2a$$
The triangle PAQ is made up of points lying on the parabola except the origin. Their centroid is
$$x=\frac{h+at_1^2+at_2^2}{3}$$
Since the value of $at^2$ is constant
$$x=\frac{3h-4a}{3}$$
And $$y=0$$ because it seemed likely. I have no idea why though
The answer is $$(\frac 23 (h-2a),0)$$
Best Answer
Normal(s) of $y^2=4ax$ at a $t$ point pass through $(h,k)$. So $$At^3+(2a-h)t-k=0$$ $t$ has three roots then $$t_1+t_2+t_3=0 ~~~(1),~~~ t_1t_2+t_2t_3+t_3t_1=(2a-h)/a~~~~(2)$$ The cenroid of the $\Delta PQR$ is $G=[a(t^2_1+t^2_2+t^2_3)/3, 2a(t_1+t_2+t_3)/3]$ From (1) and (2) $$t^2_1+t^2_2+t^3+3=(t_1+t_2+t_3)^2-2(t_1 t_2+t_2 t_3+t_3 t_1)=2(h-2a)$$ Finall the co-ordinates of centroid is $$G=[\frac{2}{3}(h-2a),0]$$