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Question

If normal at point P(6,2) on the curve xyx2y2=0 intersects the curve again at Q(a,b), then the value of 12a+b is

A
8
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B
3
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C
6
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D
12
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Solution

The correct option is C 6
xyx2y2=0 (1)
Differentiate with respect to x
y+xdydx12dydx=0
satisfies the point (6,2)
dydx=14
slope of normal =4
Equation of normal is
y2=4(x6)
y4x+22=0 (2)
From equation (1) and (2), we get
x(4x22)x2(4x22)2=0
4x231x+42=0
x=6,74
x=6 is point P, so x=74 is point Q
Thus point where normal interests the curve again is Q(74,15)
The value of 12a+b=12×7415
=6

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