How-do-you-find-a-point-on-the-curve-y-x-2-that-is-closest-to-the-point-16-1-2-

Question Number 139255 by bobhans last updated on 25/Apr/21

How do you find a point on the curve y = x^2 that is closest to the point? (16, 1/2)

Answered by john_santu last updated on 25/Apr/21

T h e t a n g e n t t o p a r a b o l a a t i t s p o i n t

(r, r^{2}) h a s e q u a t i o n y = 2 r x - r^{2}

T h e n o r m a l f o r r \neq 0 w i l l h a v e

g r a d i e n t m = - \frac{1}{2 r} a n d i t w i l l h a v e

e q u a t i o n y = - \frac{x}{2 r} + r^{2} + \frac{1}{2}

s u c h a l i n e m u s t p a s s e s t h r o u g h

a t p o i n t (16, \frac{1}{2}) w e g e t

\frac{1}{2} = - \frac{16}{2 r} + r^{2} + \frac{1}{2} t h a t s i m p l i f i e s

t o r^{3} = 8 s o r = 2 a n d t h e p o i n t

o f m i n i m u m d i s t a n c e i s (2, 4)

Answered by mr W last updated on 25/Apr/21

d i s t a n c e f r o m p o i n t (16, \frac{1}{2}) t o p o i n t

(x, y) o n t h e c u r v e y = x^{2} i s d .

D = d^{2} = {(x - 16)}^{2} + {(y - \frac{1}{2})}^{2}

D = {(x - 16)}^{2} + {(x^{2} - \frac{1}{2})}^{2}

\frac{d D}{d x} = 2 (x - 16) + 2 (x^{2} - \frac{1}{2}) (2 x) = 0

x^{3} = 8

x = 2

y = 2^{2} = 4

\Rightarrow p o i n t (2, 4) i s c l o s e s t t o (16, \frac{1}{2}) .

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