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Question

If the line x−y=−4K is a tangent to the parabola y2=8x at P, then the perpendicular distance of normal at P from (K,2K) is

A
522
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B
722
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C
922
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D
122
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Solution

The correct option is A 922
xy=4ky=x+4k
m=1,c=4k
It is a tangent to the parabola y2=8xa=84=2
c=am4k=21k=12
xy=4(12)xy+2=0
Point of contact p=(am2,2am)=(2,4)
Equation of normal at p(2,4) is y=mx2amam3
slope of tangent at p is 1
Equation of normal a p is y=x+6x+y6=0
Perpendicular distance from (k,2k)=(12,1) to x+y6=0
∣ ∣ ∣ ∣12+262∣ ∣ ∣ ∣=922

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