A designer sketches the cross-section of a reflector on graph paper, with lengths in metres…

Step 1. Substitute. $\dfrac{dy}{dx} = \dfrac{y}{x} + \sqrt{1 + \left(\dfrac{y}{x}\right)^2}$. With $y = vx$: $v + x\dfrac{dv}{dx} = v + \sqrt{1 + v^2}$. So $\dfrac{dv}{\sqrt{1 + v^2}} = \dfrac{dx}{x}$.

Step 2. Integrate. $\ln\left(v + \sqrt{1 + v^2}\right) = \ln x + c$, so $v + \sqrt{1 + v^2} = Cx$. Multiply by $x$: $y + \sqrt{x^2 + y^2} = Cx^2$.

Step 3. The constant. At $(1, 0)$: $0 + \sqrt{1} = C$, so $C = 1$.

Step 4. Solve for $y$. Write $\sqrt{x^2 + y^2} = x^2 - y$ and square: $x^2 + y^2 = x^4 - 2x^2y + y^2$. The $y^2$ cancels, so $2x^2y = x^4 - x^2$ and $y = \dfrac{x^2 - 1}{2}$.

Step 5. Evaluate. At $x = 2\ \text{m}$: $y = \dfrac{4 - 1}{2} = 1.5\ \text{m}$. At $x = 3\ \text{m}$: $y = \dfrac{9 - 1}{2} = 4\ \text{m}$.