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the length of the chord of the parabola y^2=x which is bisected at the point (2,1) is
T = S1y-0.5(x+2) = 1^2 - 2 = -12y - x - 2 = -2or, x = 2yPut in the equation,y^2 = 2yor, y = 0 , 2x = 0, 4Points where the line cuts the parabola = (0,0), (4,2)Length = sqrt(4^2 + 2^2) = sqrt(16+4) = sqrt(20) = 2*sqrt(5)Thanks & RegardsRinkoo GuptaAskIITians Faculty
T = S1
y-0.5(x+2) = 1^2 - 2 = -1
2y - x - 2 = -2
or, x = 2y
Put in the equation,
y^2 = 2y
or, y = 0 , 2
x = 0, 4
Points where the line cuts the parabola = (0,0), (4,2)
Length = sqrt(4^2 + 2^2) = sqrt(16+4) = sqrt(20) = 2*sqrt(5)
Thanks & Regards
Rinkoo Gupta
AskIITians Faculty
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