A curve passing through the point (1,1) has the porperty that the perpendicular distance of the normal at any point P on the curve from the origin is equal to the distance of P from x-axis Determine the equation of the curve.

Explanation

Equation of normal is `(dx)/(dy)(X-y) + Y-y=0`
Given that perpendicular distance form origin to the normal at P=distance of P from the x-axis.
or `|(x(dx)/(dy)+y)/(1+((dx)/(dy))^(2))|= |y|`
or `x^(2)((dx)/(dy))^(2)+y^(2)+2xy(dx)/(dy) = y^(2)+y^(2)(dx)/(dy)^(2)`
i.e., `(dx)/(dy) =0` or `(dx)/(dy) =(2xy)/(y^(2)-x^(2))`
if `(dx)/(dy)=0`, then x=c, When `x=1, y=1`, then `c=1`.
therefore `x=1` ................(1)
`(dx)/(dy) = (2xy)/(y^(2)-x^(2))` (homogeneous) ...............(2)
Putting, `x=vy` and `(dx)/(dy)=v+y(dv)/(dy)`, equation (2) transforms to
`v+y(dv)/(dy)= (2v)/(1-v^(2))`
or `y(dv)/(dx) = (2v)/(1-v^(2))-v=(2v-v+v^(2))/(1-v^(2)) = (v+v^(3))/(1-v^(2))`
or `int((1-v^(2))dv)/(v(1+v^(2))-v)= (2v-v+v^(3))/(1-v^(2))=(v+v^(3))/(1-v^(2))`
or `int((1-v^(2)dv))/(v(1+v^(2)))=int(dy)/(y)`
or `int(1/v-(2v)/(1+v^(2)))dy=int(dy)/(y)=C`
or `v/(1+v^(2))=cy`
or `x/(x^(2)+y^(2))=c`
When `x=1, y=1`, `c=1//2`.
Thus, solution is `x^(2)+y^(2)-2x=0`.
Hence, the solution are `x^(2)+y^(2)-2x=0, x-1=0`