Find the equation of the curve through (1,0) for which the y…

Find the equation of the curve through $(1,0)$ for which the $y$-intercept of the tangent line equals the distance from the point of tangency to the origin.

Step 1: Set up and solve the quadratic in $y'$

Squaring $y-xy'=\pm\sqrt{x^2+y^2}$ gives $x^2(y')^2-2xyy'-x^2=0$, i.e. $(y')^2-\tfrac{2y}{x}y'-1=0$, so

\[y'=\frac{y\pm\sqrt{x^2+y^2}}{x}.\]

Step 2: Solve the $+$ branch by $y=vx$

\[x\frac{dv}{dx}=\sqrt{1+v^2} \quad\Longrightarrow\quad v+\sqrt{1+v^2}=Kx.\]

With $y(1)=0$ (so $v(1)=0$): $K=1$. Squaring $\sqrt{1+v^2}=x-v$ gives $v=\dfrac{x^2-1}{2x}$, so

\[\boxed{\,y=vx=\frac{x^2-1}{2}.}\]

Step 3: Verify

With $y=\tfrac{x^2-1}{2}$, $y'=x$, so $y-xy'=-\tfrac{x^2+1}{2}$, whose absolute value $\tfrac{x^2+1}{2}$ indeed satisfies $\left(\tfrac{x^2+1}{2}\right)^2=x^2+\left(\tfrac{x^2-1}{2}\right)^2$ (both sides equal $\tfrac{x^4+2x^2+1}{4}$). By the symmetry $y\mapsto-y$, the curve $y=-\tfrac{x^2-1}{2}$ also passes through $(1,0)$ and satisfies the condition.