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Question

The equation of the curve passing through the origin if the middle point of the segment of its normal form any point of the curve to the x-axis, lies on the parabola 2y2=x

A
y2=2x+1e2x
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B
y2=2x+1+e2x
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C
y2=2x+e2x
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D
y2=x+1e4x
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Solution

The correct option is D y2=2x+1e2x
Equation of normal at any point P(x,y) is
dydx(Yy)+(Xx)=0
This meets the x-axis at A(x+ydydx,0).
Mid-point of AP is (x+12ydydx,y2) which lies on the parabola 2y2=x.
2×y24=x+12ydydxy2=2x+ydydx
Putting y2=t
we get dtdx2t=4x
I.F.=e2dx=e2x
Therefore, the solution is given by
te2x=4xe2xdx+cte2x=4[12e2dx]+cy2e2x=2xe2x+e2x+c
Since, curve passes through (0,0),c=1. Therefore,
y2=2x+1e2x is the equation of the required curve.

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