A curve passing through the point (1, 1) has the property 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. Then length of intercept of equation of the curve on the line x-2y=0 is

Let P (x, y) be any point on the curve y = f(x). Then the equation of the normal at P is.
$Y-y=-\frac{1}{(dy/dx)}(X-x)$ Or $X+Y\frac{dy}{dx}-(Y\frac{dy}{dx}+x)=0\mathrm{...}\left(i\right)$  
It is given that distance of E.q. (i) from origin = Distance from x-axis (i.e.y) 
i.e. $\left|\frac{0-(y\frac{dy}{dx}+x)}{\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}}\right|=y$  
$\Rightarrow {(y\frac{dy}{dx}+x)}^2=y^2[1+{\left(\frac{dy}{dx}\right)}^2]$ 
$\Rightarrow x^2+2xy\frac{dy}{dx}=y^2$ Or $\frac{dy}{dx}=\frac{y^2-x^2}{2xy}$  
Which is homogeneous differential equation and we can solve by homogeneous or by total differential.
Here, using total differential, $2xydy-y^{2}dx=-x^{2}dx$ 
$\Rightarrow \frac{xd\left(y^2\right)-y^{2}dx}{x^2}=-dx\Rightarrow d\left(\frac{y^2}{x}\right)=-dx$
Integrating both the sides, we get $\Rightarrow \frac{y^2}{x}=-x+C$
It passes through (1, 1) $\Rightarrow C=2$  
$\therefore \frac{y^2}{x}=-x+2or\text{\hspace{0.17em}\hspace{0.17em}}{y}^2=-x^2+2x$ 
$\Rightarrow x^2+y^2-2x=0,$ is required equation of curve.