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# Find the equation of the curve passing through the orgin if the middle point of the segment of its normal from any point of the curve to the x-axis lies on the parabola 2y2=x.

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Solution

## Find the equation of curve passing through the origin if the middle point of its normal from any point of the curve to the x−axis lies on the parabola 2y2=xEquation of normal at point p(x,y) is y−x=dydx(X−x)Normal meets at x− axis at Q(X,0)y−x=dydx(X−x) X=ydydx+xQ=(ydydx+x,0)Let middle point of PQ is R which is ⎛⎜ ⎜ ⎜⎝x+x+ydydx2,y+02⎞⎟ ⎟ ⎟⎠=⎛⎜ ⎜ ⎜⎝2x+ydydx2,y2⎞⎟ ⎟ ⎟⎠R lies on parabola 2y2=x Let y2=t..............(1)2ydydx=dtt=2x+12dtdxdtdx−2t=−4xIF=e∫(−2)dx=e−2xt(IF)=∫(−4x)(IF)dx+Cte−2x=−4x−2e−2x−∫−4x−2e−2x+Cte−2x=2xe−2x+e−2x+Ct=2x++1+ce2xFrom (1)y2=2x+1+ce2x If this passes through origin as given in QC=−1Therefore the equation of required parabola is y2=2x+1−e2x

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