If the normal at the point P(ap2,2ap) meets the parabola at Q(aq2,2aq) such that the lines joining the origin to P and Q are at right angle, then
A
p2=2
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B
pq−4=0
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C
pq+4=0
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D
p2=4
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Solution
The correct option is Cpq+4=0 Equation of the normal at (ap2,2ap) to y2=4ax is y+px=2ap+ap3
It passes through Q(aq2,2aq), so ∴2aq+aq2p=2ap+ap3⇒(p−q)(2+p2+pq)=0[∵a≠0]∴2+p2+pq=0⋯(1)[∵p≠q]