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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 $$(\frac 23 (h-2a),0)$$

Aditya
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1 Answers1

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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 $$G=[\frac{2}{3}(h-2a),0]$$

Z Ahmed
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  • Can it be solved the way I did it? I know it isn’t the right way, but I almost got it all the way through to the end – Aditya Mar 02 '20 at 12:55
  • You cannot put $k=0$, it is non-zero and the $t$ eq. is a cubic. Rest you can see my solution.. Cheers – Z Ahmed Mar 02 '20 at 14:03
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    But no where it’s said that k is non zero. Also, I eliminated one $t$ from the cubic equation, since it was zero – Aditya Mar 02 '20 at 14:15
  • @Aditya Yes, you have a point when you put $k=0$ you get $t=0=t_3$ Use $G$ point as I wrote, do this to get correct answer.. – Z Ahmed Mar 02 '20 at 15:37