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Question

Consider a parabola y2=4x and F be its focus. Let A1(x1,y1),A2(x2,y2),A3(x3,y3),...,An(xn,yn) be n points on the parabola such that xk,yk>0 and OFAk=kπ2n. If limn1nnk=1FAk=Pπ, then the value of P is

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Solution



Slope of FAk is
yk0xk1=tanθk
xk=1yktanθk

We know that y2k=4xk
y2k=4(1yktanθk)
y2k+(4cotθk) yk4=0
yk=4cotθk±16cot2θk+162
=4cotθk+4 cosec θk2,4cotθk4 cosec θk2
=2(1cosθksinθk),2(1+cosθksinθk)

yk=2tanθk2,2cotθk2
We know θk=kπ2nπ2 and (xk,yk) is in first quadrant so,
yk=2tanθk2
4tan2θk2=4xk
xk=tan2θk2
FAk=(xk1)2+(yk)2FAk=(xk1)2+4xkFAk=(xk+1)FAk=1+tan2θk2FAk=sec2θk2=sec2kπ4n

limn1nnk=1FAk=10sec2(πx4)dx
=4π[tan(πx4)]10=4π=Pπ

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