Find the point p on the parabola y = x2 closest to the point (57, 0).
Answers
Answered by
2
y = x² is a parabola ,
Let ( t , t²) is a point on the parabola which is closest to the point ( 57, 0) .
distance ( S) = √{ (t - 57)² + (t²)² }
S = √{ t² + 57² -114t + t⁴}
differentiate with respect to t
dS/dt = 1/2S { 2t + 0 - 114 + 4t³} = 0
2t - 114 + 4t³ = 0
2t³ + t - 57 = 0
2t³ - 6t² + 6t² - 18t + 19t - 57 = 0
2t²( t - 3) + 6t( t -3) + 19(t -3) = 0
t = 3 and 2t² + 6t + 19 ≠ 0
again, differentiate wrt t
d²S/dt² > 0 at t = 3
so, at t = 3 S is minimum
hence,
( 3 , 9) is unknown points .
Let ( t , t²) is a point on the parabola which is closest to the point ( 57, 0) .
distance ( S) = √{ (t - 57)² + (t²)² }
S = √{ t² + 57² -114t + t⁴}
differentiate with respect to t
dS/dt = 1/2S { 2t + 0 - 114 + 4t³} = 0
2t - 114 + 4t³ = 0
2t³ + t - 57 = 0
2t³ - 6t² + 6t² - 18t + 19t - 57 = 0
2t²( t - 3) + 6t( t -3) + 19(t -3) = 0
t = 3 and 2t² + 6t + 19 ≠ 0
again, differentiate wrt t
d²S/dt² > 0 at t = 3
so, at t = 3 S is minimum
hence,
( 3 , 9) is unknown points .
Similar questions