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Find the equation of the curve passing t...

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`.

Text Solution

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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.
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