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Question

The abscissa of the point on the curve yx=(x+a)2 at which the normal cuts off numerically equal intercepts on the coordinate axes is

A
2a2
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B
a2
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C
2a
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D
a
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Solution

The correct option is C 2a
For normal to cut numerically equal intercepts there are few possibilities.
If it cuts intercepts of (1,1), (2,2) etc then slope =1. If it cuts intercepts of (1,1), (2,2) etc. then slope = 1.

Therefore, for normal to cut numerically equal intercepts slope must be =±1

yx=(x+a)2
On differentiating the above equation w.r.t x, we have-
y+xy=2(x+a)
y=2(x+a)yx= slope of tangent
Let Mnormal be the slope of normal =x2(x+a)y
=x2(x+a)(x+a)2x [yx=(x+a)2]
=x22x(x+a)(x+a)2=±1
For Mnormal=1
2x(x+a)(x+a)2=x2
2x2x2a2=x2
x=12a
For Mnormal=1
2x2+x2+a2=x2
No solution
x=12a

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