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P and Q are 2 variable points with parameters p and q respectively on the parabola 8y= #x^2# such that OP is perpendicular to OQ where O is the origin. Given equation of PQ is y=#x/(2sqrt2)(p+q)-2pq# how do you show that chord PQ passes through a fixed point?

1 Answer
P dilip_k
Nov 25, 2017

Given that P and Q are 2 variable points with parameters p and q respectively on the parabola #8y=x^2#, we can write the coordinates of two variable points are

#P->(2sqrt2p,p^2)#

#Q->(2sqrt2q,q^2)#

Origin #O->(0,0)#

As OP is perpendicular to OQ , the product of their slope #=-1#

#=>p^2/(2sqrt2p)xxq^2/(2sqrt2q)=-1#

#=>pq=-8#

Now given equation of PQ ,#y=x/(2sqrt2)(p+q)-2pq# becomes

#y=x/(2sqrt2)(p+q)-2(-8)#

#or,y=x/(2sqrt2)(p+q)+16#

Inserting #x=0# we get #y=16#

So the chord PQ passes through the fixed point #(0,16)#

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