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Question

# A curve is such that the mid point of the segment of its normal intercepted between the point from where it is drawn and the point where the normal meets the x−axis lies on the parabola 2y2=x. If the curve passes through origin, then the curve is

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

## The correct option is A y2=2x+1−e2xEquation of normal at P(x,y) is Y−y=−dxdy(X−x) 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 Put y2=t⇒ydydx=12dtdx We get dtdx−2t=−4x (which is linear) I.F.=e−2∫dx=e−2x ∴ Solution is te−2x=−4∫xe−2xdx+c On integrating by parts, we have: te−2x=−4[xe−2x−2−∫−12e−2xdx]+c ⇒y2e−2x=2xe−2x+e−2x+c Since the curve passes through origin, the value of c=−1 ∴y2=2x+1−e2x is the equation of the required curve.

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