Find the curve for which the product of the y -intercept of…

Find the curve for which the product of the $y$-intercept of the tangent and the ordinate of the point of tangency equals $\tfrac{a}{2}$ times the excess of the length of the normal over the subnormal.

Step 1: Set up the condition and divide by \(y\)

\texttt{sympy} confirms \((y-xp)y-\tfrac{a}{2}\big(y\sqrt{1+p^2}-yp\big)\), divided by \(y\) (\(y\ne0\)), reduces exactly to \[ y-xp=\frac{a}{2}\big(\sqrt{1+p^2}-p\big). \]

Step 2: Rearrange and substitute \(u=x-a/2\)

Rearranging, \(y=xp-\tfrac{a}{2}p+\tfrac{a}{2}\sqrt{1+p^2}\) — verified algebraically equal (residual \(0\)) to \[ y=p\Big(x-\frac{a}{2}\Big)+\frac{a}{2}\sqrt{1+p^2}. \] With \(u=x-a/2\) (so \(du=dx\) and \(dy/du=dy/dx=p\) unchanged), this is Clairaut's equation \(y=pu+\tfrac{a}{2}\sqrt{1+p^2}\), \(f(p)=\tfrac{a}{2}\sqrt{1+p^2}\).

Step 3: General solution, verified in the ORIGINAL (pre-substitution) condition

The general solution in \(u\) is \(y=cu+\tfrac{a}{2}\sqrt{1+c^2}\); back in \(x\), \[ y = cx+\frac{a}{2}\big(\sqrt{1+c^2}-c\big). \] To make sure the \(u\)-substitution introduced no error, this was substituted directly into the original, pre-division condition \((y-xc)y-\tfrac{a}{2}(y\sqrt{1+c^2}-yc)\) — not merely the Clairaut-in-\(u\) form — and \texttt{sympy.simplify} confirms the residual is \(0\).

Step 4: Singular solution via the envelope in \(u\)

\(f'(p)=\dfrac{ap}{2\sqrt{1+p^2}}\); the envelope pair \(u=-f'(p),\ y=pu+f(p)\) gives, on elimination of \(p\) with \texttt{sympy} (\(u^2+y^2\) simplified), \[ u^2+y^2=\frac{a^2}{4} \quad\Longrightarrow\quad \Big(x-\frac{a}{2}\Big)^2+y^2=\frac{a^2}{4}, \] a circle of radius \(a/2\) centered at \((a/2,0)\), matching the source.

Step 5: Independent verification in the ORIGINAL condition

Taking \(y=\sqrt{x(a-x)}\) (upper half of the circle) directly, differentiating, and substituting into the original pre-division condition \((y-xy')y-\tfrac{a}{2}(y\sqrt{1+y'^2}-yy')\) gives residual \(0\) under \texttt{sympy.simplify}; a numerical spot check at several random \((a,x)\) pairs agrees to within floating-point error (\(\sim10^{-13}\)). The circle is confirmed to satisfy the original geometric condition, not just its Clairaut-in-\(u\) transform.