Find the equation of a curve passing through the point (1.1) if the perpendicular distance of the origin from the normal at any point P(x, y) of the curve is equal to the distance of P from the x-axis.

Answer

`(x^(2)+y^(2) = 2x)`

Explanation

Equation of normal at point (x,y) is
`Y-y=-dx/dy (X-x) ...(i)`
Distance of perpendicular from the origin to Eq. (i)
`= (abs(y+dx/dy cdotx))/sqrt(1+(dx/dy)^(2))`
Also, distance between P and X-axis is `abs(y)` .
`therefore = (abs(y+dx/dy cdotx))/sqrt(1+(dx/dy)^(2))=abs(y)`
`rArr y^(2)+dx/dy cdot x^(2) +2xy dx/dy=y^(2)[1+(dx/dy)^(2)]`
`rArr (dx/dy)^(2) (x^(2)-y^(2))+2xy dx/dy=0`
`rArr dx/dy [(dx/dy) (x^(2)-y^(2))+2xy]=0`
`rArr dx/dy=0 or dy/dx=(y^(2)-x^(2))/(2xy) `
But `dx/dy=0`
`rArr x=c,` wher c is a constant.
Since, curve passes through (1,1), we get the equation
of the curve as `x=1.`
The equation `dy/dx=(y^(2)-x^(2))/(2xy)` is a homogeneous equation.
Put `y=vx rArr dy/dx=v+x(dv)/dx`
`v+x(dv)/dx= (v^(2) x^(2)-x^(2))/(2x^(2)v)`
`rArr x(dv)/dx=(v^(2)-1)/(2v)-v=(v^(2)-1-2v^(2))/(2v)=-(v^(2)+1)/(2v)`
`rArr (-2v)/(v^(2)+1)dv=dx/x`
`c_(1)-log (v^(2)+1)=logabs(x)`
` =log abs(x) (v^(2)+1)=c_(1) rArr abs(x)(y^(2)/x^(2)+1)=e^(c_(1))`
`rArr x^(2)+y^(2)=pme^(c_(1)) x or x^(2)+y^(2)=pm e^(c)` is passinf through
(1,1).
`therefore 1+1=pm e^(c).1`
`rArr pm e^(c)=2`
Hence, required curve is `x^(2)+y^(2)=2x.`