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Question

Find the point on the curve x2=8y which is nearest to the point (2,4).

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Solution

Let x, y be nearest to the point 2, 4. Then,x2=8yy=x28 ...1d2=x-22+y-42 Using distance formulaNow,Z=d2=x-22+y-42Z=x-22+x28-42 From eq. 1Z=x2+4-4x+x464+16-x2dZdy=-4+4x364For maximum or minimum values of Z, we must havedZdy=0-4+4x364=0x316=4x3=64x=4Substituting the value of x in eq. 1, we gety=2Now,d2Zdy2=12x264d2Zdy2=3>0So, the nearest point is 4, 2.

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