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Question

The line L:y=mxb touches the parabola P:y=ax2 where a and m are positive real constant and b is real constant, at the point T. Let Q be the point of intersection of the line L and yaxis such that TQ=1. If M is the maximum value of the area of the region surrounded by P,L and yaxis, then the value of 1M is

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Solution

Given : P:y=ax2
Let T=(t2a,t24a)
Now,
dydxT=t=m
So, the equation of L becomes
y=txb
T lies on the line, so
t24a=t22abb=t24ay=txt24aQ=(0,b)=(0,t24a)


Now,
TQ=1t24a2+t44a2=1t2(t2+1)=4a2(1)

Therefore,
A=12×(t24a+t24a)×t2a1at24a0y dyA=t38a223a(t24a)3/2A=t324a2
Using equation (1), we get
A=t36t2(t2+1)A=16(t+1t)
So,
M=16×21M=12

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