Find the equation of the curve passing through the origin if the middle point of the segment of its normal from any point of the curve to the x-axis lies on the parabola `2y^2=x`

Explanation

Equation of normal at any point `P(x,y)`, is
`(dy)/(dx)(Y-y)+(X-x)=0`
This meets the x-axis at `A(x+y(dy)/(dx),0)`
Mid-point of AP is `(x+1/2y(dy)/(dx),y/2)` which lies on the parabola `2y^(2)=x`. Therefore, `2 xx y^(2)/4=x+1/2y(dy)/(dx)` or `y^(2)=2x+y(dy)/(dx)`
so that `2y(dy)/(dx) = (dt)/(dx)` ,
we get `(dt)/(dx)-2t=-4x` (linear)
I.F. `e^(-2)int(dx)=e^(-2x)`
Therefore, solution is given by
`te^(-2x)=-4intxe^(-2x)dx+c`
`=-4[-1/2xe^(-2x)-int-1/2e^(-2x)1dx]+c`
`y^(2)e^(-2x)=2xe^(-2x) + e^(-2x)+c`
Since, the curve passes through (0,0), `c=-1`. Therefore,
`y^(2)e^(-2x)+e^(-2x)-1`
`y^(2)=2x+1-e^(2x)` is the equation of the required curve.