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Question

Let the tangent at point P(8,4) to the curve y3=x2 meets the curve again at Q and PQ=a10. Then the value of a is

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Solution

Given curve : y3=x2
Let point Q=(t3,t2)
For slope of tangent at P:3y2dydx=2x
So, slope of tangent at P(8,4)= slope of PQ
(dydx)(8,4)=t24t38
13=t24t38
13=(t2)(t+2)(t2)(t2+2t+4)
t2t2=0
t=1,2; (t2)
So, Q(1,1)
PQ=81+9=310
a=3

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