1. **State the problem:** Find the equation of a curve passing through the point (1, 1) such that the perpendicular distance of the origin from the normal at any point $P(x,y)$ on the curve equals the distance of $P$ from the x-axis. 2. **Set up variables and expressions:** Let the curve be $y = f(x)$ and its derivative $y' = \frac{dy}{dx} = m$. 3. **Equation of the normal at $P(x,y)$:** The slope of the tangent is $m$, so the slope of the normal is $-\frac{1}{m}$. 4. The normal line at $P$ is: $$y - y_1 = -\frac{1}{m}(x - x_1)$$ where $P = (x_1, y_1) = (x, y)$. 5. **Distance of origin $(0,0)$ from the normal line:** The line can be rewritten as: $$m(y - y_1) + (x - x_1) = 0$$ or $$m y - m y_1 + x - x_1 = 0$$ 6. The distance $d$ from origin to this line is: $$d = \frac{|m \cdot 0 - m y_1 + 0 - x_1|}{\sqrt{m^2 + 1}} = \frac{| - m y - x|}{\sqrt{m^2 + 1}} = \frac{|x + m y|}{\sqrt{m^2 + 1}}$$ 7. **Distance of $P$ from x-axis:** This is simply the absolute value of the y-coordinate: $$|y|$$ 8. **Given condition:** The distance of origin from normal equals distance of $P$ from x-axis: $$\frac{|x + m y|}{\sqrt{m^2 + 1}} = |y|$$ 9. Since $y$ can be positive or negative, square both sides to avoid absolute values: $$\frac{(x + m y)^2}{m^2 + 1} = y^2$$ 10. Multiply both sides by $m^2 + 1$: $$(x + m y)^2 = y^2 (m^2 + 1)$$ 11. Expand left side: $$x^2 + 2 x m y + m^2 y^2 = y^2 m^2 + y^2$$ 12. Cancel $m^2 y^2$ from both sides: $$x^2 + 2 x m y = y^2$$ 13. Recall $m = \frac{dy}{dx}$, rewrite: $$x^2 + 2 x y \frac{dy}{dx} = y^2$$ 14. Rearrange to isolate $\frac{dy}{dx}$: $$2 x y \frac{dy}{dx} = y^2 - x^2$$ $$\frac{dy}{dx} = \frac{y^2 - x^2}{2 x y}$$ 15. This is a separable differential equation. Rewrite: $$\frac{dy}{dx} = \frac{y^2}{2 x y} - \frac{x^2}{2 x y} = \frac{y}{2 x} - \frac{x}{2 y}$$ 16. Multiply both sides by $2 y$: $$2 y \frac{dy}{dx} = y^2 / x - x$$ 17. Alternatively, keep original form and separate variables: $$\frac{dy}{dx} = \frac{y^2 - x^2}{2 x y}$$ 18. Rewrite as: $$2 x y dy = (y^2 - x^2) dx$$ 19. Rearrange: $$2 x y dy - (y^2 - x^2) dx = 0$$ 20. Write as: $$2 x y dy - y^2 dx + x^2 dx = 0$$ 21. Group terms: $$x^2 dx - y^2 dx + 2 x y dy = 0$$ 22. Check if exact differential: Let $M = x^2 - y^2$, $N = 2 x y$. Calculate partial derivatives: $$\frac{\partial M}{\partial y} = -2 y$$ $$\frac{\partial N}{\partial x} = 2 y$$ Since $\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$, not exact. 23. Try integrating factor $\mu = \frac{1}{x^2}$: Multiply entire equation by $\frac{1}{x^2}$: $$dx - \frac{y^2}{x^2} dx + \frac{2 y}{x} dy = 0$$ Now $M = 1 - \frac{y^2}{x^2}$, $N = \frac{2 y}{x}$. Calculate partial derivatives: $$\frac{\partial M}{\partial y} = -\frac{2 y}{x^2}$$ $$\frac{\partial N}{\partial x} = -\frac{2 y}{x^2}$$ Equal, so equation is exact after multiplying by $\frac{1}{x^2}$. 24. Find potential function $\psi(x,y)$ such that: $$\frac{\partial \psi}{\partial x} = M = 1 - \frac{y^2}{x^2}$$ Integrate w.r.t. $x$: $$\psi = \int \left(1 - \frac{y^2}{x^2}\right) dx = x + y^2 \frac{1}{x} + h(y) = x + \frac{y^2}{x} + h(y)$$ 25. Differentiate $\psi$ w.r.t. $y$: $$\frac{\partial \psi}{\partial y} = \frac{2 y}{x} + h'(y)$$ Set equal to $N = \frac{2 y}{x}$: $$\frac{2 y}{x} + h'(y) = \frac{2 y}{x} \implies h'(y) = 0$$ So $h(y)$ is constant. 26. The implicit solution is: $$\psi(x,y) = C \implies x + \frac{y^2}{x} = C$$ 27. Multiply both sides by $x$: $$x^2 + y^2 = C x$$ 28. Use initial condition $(1,1)$: $$1^2 + 1^2 = C \cdot 1 \implies 2 = C$$ 29. Final equation of the curve: $$x^2 + y^2 = 2 x$$ 30. Rewrite as: $$(x - 1)^2 + y^2 = 1$$ This is a circle centered at $(1,0)$ with radius $1$. **Answer:** The curve is the circle $$ (x - 1)^2 + y^2 = 1 $$