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Question

Consider the curve x=13t2,y=t3t3, Let P(-2, 2) be a point on the curve. Let the tangent at P
cuts the curve again at Q. Then answer the following
The Point Q will be

A
(2,2)
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B
(12,29)
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C
(23,29)
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D
(2,1)
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Solution

The correct option is B (12,29)
x=13t2,y=t3t3P=(2,2)t=1dydx=19t26t,(dydx)t=1=43
Equ of tangent at P is y2=43(x+2)
Q also lies on the above tangent
t3t32=43(13t2+2)3t3+4t2t2=0
(3t2)(t+1)2=0t=23for Q
Q=(13,29)

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