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Question

If the tangent at (1, 1) on y2 = x(2 x)2 meets the curve again at P, then P is

A
(4, 4)
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B
(-1,2)
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C
(94, 38)
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D
(0,0)
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Solution

The correct option is C (94, 38)
2y dydx=(2x)2 - 2x(2 - x), so dydx|(1,1)=12[12]=12
Therefore, the equation of tangent at (1, 1) is y - 1 = -12(x1)
2y - 2 = -x + 1
y = x+32
The intersection of the tangent and the curve is given by (14)(x+3)2=x(4+x24x)
x26x+9=16x+4x316x2
4x317x2+22x9=0
(x1)(4x213x+9)=0
(x1)2(4x9)=0
Since x = 1 is already the point of tangency x = 94 and y2 = 94(294)2=964
Thus the required point is (94,38)

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