Weird coordinates

Calculus Level 4

P P is a point such that P P 1 2 + P P 2 2 + . . . . . P P 5 2 = k PP_{1}^{2}+PP_{2}^{2}+.....PP_{5}^{2}=k where P r P_{r} is point with coordinates ( r , r 2 ) (r,r^2) .The least value of k k for which such point P exists is

384 N o n e None 374 386 378

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1 solution

Tanishq Varshney
Feb 8, 2015

k = ( x x 1 ) 2 + ( y y 1 ) 2 + ( x x 2 ) 2 + ( y y 2 ) 2 . . . . . . k=(x-x_1)^{2}+(y-y_1)^{2}+(x-x_2)^{2}+(y-y_2)^{2}......

Differentiating by first keeping y constant and then x

d k d x = 2 x 2 x 1 + 2 x 2 x 2 + . . . + 2 x 2 x n \frac{dk}{dx}=2x-2x_{1}+2x-2x_{2}+...+2x-2x_{n}

and

d k d y = 2 y 2 y 1 + 2 y 2 y 2 . . . . . . . + 2 y 2 y n \frac{dk}{dy}=2y-2y_{1}+2y-2y_{2}.......+2y-2y_{n}

equating

d k d x = 0 \frac{dk}{dx}=0 and d k d y = 0 \frac{dk}{dy}=0

we get x = x 1 + x 2 + . . . . x n n x=\frac{x_{1}+x_{2}+....x_{n}}{n}

y = y 1 + y 2 + . . . . y n n y=\frac{y_{1}+y_{2}+....y_{n}}{n}

Using second derivative test

k k is minimum for the values of x x and y y .

Thus, minimum value of k k is sum of squares of distances from c e n t r o i d centroid

(denoted as G G )

G ( 1 + 2 + 3 + 4 + 5 5 , 1 2 + 2 2 + 3 2 + 4 2 + 5 2 5 ) G(\frac{1+2+3+4+5}{5},\frac{1^{2}+2^{2}+3^{2}+4^{2}+5^{2}}{5})

G ( 3 , 11 ) G(3,11)

k = 2 2 + 1 0 2 + 1 2 + 7 2 + 0 2 + 2 2 + 1 2 + 5 2 + 2 2 + 1 4 2 k=2^{2}+10^{2}+1^{2}+7^{2}+0^{2}+2^{2}+1^{2}+5^{2}+2^{2}+14^{2}

k = 384 \boxed {k=384}

plz upvote and follow

Hey Tanishq how much are you getting in open test(mains)(If given)

mudit bansal - 6 years, 4 months ago

Are you in FIITJEE? This question was asked in the open test this year.

Purushottam Abhisheikh - 6 years, 4 months ago

The question does not mention that P cannot have coordinates of the form (r, r*r). So, if you take P=(3,9) it has smaller value of k = 188

Shobha Bagai - 2 years, 4 months ago

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