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Question

Find the shortest distance between the curve y2=x3 and 9x2+9y2−30y+16=0.?

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Solution

Note that 9x2+9y2−30y+16=0 represents the circle (x−0)2+(y−53)2=12 centered at (0,53) with radius 1.

The minimum distance between the circle and the second curve y2=x3.is the minimum distance of the second curve from the center of the circle minus one

This minimum distance is unaltered if the origin is shifted to the point(0,−53) (by parallel shifting).

By this shifting the equation of the circle reduces to x2+y2=1 (the unit circle) whose center is the origin (0,0) and the equation of the second curve becomes (y+53)2=x3.

Hence a typical point of these cond equation can be written in terms of a parameter t as (x,y)(t2,t3−53)

Therefore,the square of the distance of the point (t2,t3−53) to the origin is s(t)=t4+(t3−53)2=t4+t6−103t3+259

The minimum value of s corresponds to there solution of the equation dsdt=0

t2.(3t3+2t−5)=0 which gives two real solutions t=0,t=1.

The corresponding points in these cond curve are(0,−53)and(1,−23).

The corresponding values of s are 259and139
The minimum distance from the origin to these cond curve is √133.

Hence the minimum distance between the given circle and these cond curve is √133−1=0.20185 approximately.

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