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Question

The minimum distance between the curves x2=4y and x2+y2+18x+12y+81=0

A
72
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B
726
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C
72+6
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D
8
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Solution

The correct option is B 726
Given, circle is (x+9)2+(y+6)2=62 and parabola is x2=4y


The minimum distance occurs along the common normal of the two curves.

Equation of the normal will be x+ty=2t+t3
Since, it's passes through (9,6)
96t=2t+t3t3+8t+9=0(t+1)(t2t+8)=0t=1

So, the nearest point of the parabola is (2t,t2)(2,1)

So, the distance is 72+726=726

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