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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 A y2=2x+1−e2xEquation of normal at any point P(x,y) is dydx(Y−y)+(X−x)=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+12ydydx⇒y2=2x+ydydx Putting y2=t we get dtdx−2t=−4x I.F.=e−2∫dx=e−2x Therefore, the solution is given by te−2x=−4∫xe−2xdx+c⇒te−2x=−4[−12e−2dx]+c⇒y2e−2x=2xe−2x+e−2x+c Since, curve passes through (0,0),c=−1. Therefore, y2=2x+1−e2x is the equation of the required curve.

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