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find the point on the curve y^2=4x which is nearest to the point (2,-8)

satish , 7 Years ago
Grade 12
anser 1 Answers
Arun

Last Activity: 7 Years ago

Let (t^2,2t) is a point in given curve which is nearest from (2,-8)
S=distance between given points =root {(t^2-2)^2+(2t+8)^2}
S^2=(t^2-2)^2+(2t+8)^2
differentiate w.r.t t 
2sds/dt=2 (t^2-2) 2t +2 (2t+8)2
ds/dt=0
t^3-2t+2t+8=0
t=-2 
hence unknown point is (4,-4)
 

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