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

$$\begin{gathered} \text{ Equation of normal at point }(\mathrm{x},\mathrm{y})\text{ is } \\ Y-y=-\frac{dy}{dx}(X-x) \\ \text{ Distance of perpendicular from the origin to Eq.(1) } \\ =\frac{\left|y+\frac{dx}{dy}\cdot x\right|}{\sqrt{1+{\left(\frac{dx}{dy}\right)}^2}} \\ \therefore \frac{\left|\mathrm{y}+\frac{\mathrm{dx}}{\mathrm{dy}}\cdot x\right|}{\sqrt{1+{\left(\frac{\mathrm{dx}}{\mathrm{dy}}\right)}^2}}=\left|y\right|=\text{ distance between }\mathrm{P}\text{ and }\mathrm{x}\text{-axis } \end{gathered}$$
$\begin{array}{l}\Rightarrow y^2+\frac{dx}{dy}\cdot x^2+2xy\frac{dx}{dy}=y^2\left[1+{\left(\frac{dx}{dy}\right)}^2\right]\\ \Rightarrow {\left(\frac{dx}{dy}\right)}^2\left(x^2+y^2\right)+2xy\frac{dx}{dy}=0\\ \Rightarrow \frac{dx}{dy}\left[\left(\frac{dx}{dy}\right)\left(x^2-y^2\right)+2xy\right]=0\\ \Rightarrow \frac{dx}{dy}=0\frac{dy}{dx}=\frac{y^2-x^2}{2xy}\end{array}$
$\frac{dx}{dy}=0 \Rightarrow x=cons\tan t.$

$$\begin{gathered} \text{ Since, curve passes through }(1,1)\text{, we get the equation of the cure as }x=1\text{. } \\ \text{ The equation }\frac{dy}{dx}=\frac{y^2-x^2}{2xy}\text{ is a homogeneous equation. } \\ \text{ Substitute }\mathrm{y}=\mathrm{vx}\Rightarrow \frac{\mathrm{dv}}{\mathrm{dx}}=\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{{\mathrm{v}}^2-1-{\mathrm{x}}^2}{2{\mathrm{x}}^{2}v} \\ \begin{array}{l}\Rightarrow x\frac{dv}{dx}=\frac{v^2-1-x^2}{2x^{2}v}=-\frac{v^2+1}{2v}\\ \Rightarrow \frac{-2v}{v^2+1}dv=\frac{dx}{x}\end{array} \end{gathered}$$

$$\begin{gathered} \begin{array}{l}\Rightarrow {\mathrm{c}}_1-\log \left({\mathrm{v}}^2+1\right)=\log \left|\mathrm{x}\right|\\ \Rightarrow \log \left|\mathrm{x}\right|\left({\mathrm{v}}^2+1\right)={\mathrm{c}}_1\Rightarrow \left|\mathrm{x}\right|\left(\frac{{\mathrm{y}}^2}{{\mathrm{x}}^2}+1\right)={\mathrm{e}}^{{\mathrm{c}}_1} \\ \Rightarrow x^2+y^2=\pm e^{c_1}x\text{ or }{x}^2+y^2=\pm e^{c^c}x\text{ is passing through }(1,1)\text{. } \\ \therefore 1+1=\pm e^c.1\Rightarrow \pm e^c=2\Rightarrow e^c=2 \\ \text{ Hence, required curve is }{x}^2+y^2=2x\text{. }\end{array} \end{gathered}$$