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Question

The normals at three points P,Q,R of the parabola y2=4ax meet in (h,k). Prove that the centroid of triangle PQR on axis at distance 23(h2a) from the vertex.

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Solution

a] Equation of normals to y2=4ax is
y=mx2amam3
It passes through (h,k)
k=mh2amam3
m1+m2+m3=0;m1m2+m2m3+m3m1=2ana
Let P,Q,R be beet of there normal's
P(am22am)Q(am222am2)R(am232am3)
¯x=a3(m21+m22+m23)=a3[a22(20ha)]
=23(h2a)
¯y=2a3(m1+m2+m3)=0 Hence, proved

1204229_1296209_ans_9eadfc1c39f34b13b78661583d49e042.jpg

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