Find all curves y=y(x) such that, for the tangent and normal…

Find all curves $y=y(x)$ such that, for the tangent and normal drawn at an arbitrary point $M=(x,y)$ meeting the $x$-axis at $T$ and $N$ respectively, $OM^2=OT\cdot ON$ (O the origin).

Step 1: Tangent/normal $x$-intercepts

Tangent at $M=(x,y)$: $Y-y=p(X-x)$, meets $Y=0$ at $T=\left(x-\dfrac{y}{p},0\right)$. Normal: $Y-y=-\dfrac1p(X-x)$, meets $Y=0$ at $N=(x+yp,0)$.

Step 2: Expand $OT\cdot ON$ symbolically

\texttt{sympy.expand}$\left[\left(x-\dfrac{y}{p}\right)(x+yp)\right]$ gives exactly \[x^2+xy\left(p-\frac1p\right)-y^2,\] confirming the source's expansion.

Step 3: Impose $OM^2=OT\cdot ON$ and simplify

Setting $x^2+y^2=x^2+xy(p-1/p)-y^2$ gives $2y^2=xy(p-1/p)$. Multiplying by $p$ and solving for $x$ with \texttt{sympy.solve} gives exactly \[x=\frac{2yp}{p^2-1},\qquad\text{i.e.}\qquad x(p^2-1)=2yp,\] matching the source's relation — this algebra is confirmed correct.

Step 4: The source's claimed line-family solution is FALSE

The source claims $y=kx$ (constant slope $p=k$) solves the relation for a suitable real $k$. Substituting $y=kx$, $p=k$ into $2yp=x(p^2-1)$ gives $2k^2x=x(k^2-1)$ for all $x$, i.e. $2k^2=k^2-1\Rightarrow k^2=-1$. \texttt{sympy.solve} confirms this has \emph{no real solutions}. The source's claimed "general solution" is therefore rejected outright.

Step 5: Correct approach — treat as $x=y\,h(p)$

Write $h(p)=\dfrac{2p}{p^2-1}$, so $x=y\,h(p)$. Using $dx/dy=1/p$ and differentiating symbolically (treating $p=p(y)$): \[\frac1p=h(p)+y\,h'(p)\frac{dp}{dy}\ \Longrightarrow\ \frac{dy}{y}=\frac{h'(p)}{1/p-h(p)}\,dp.\] \texttt{sympy} computes $h'(p)=\dfrac{-2(p^2+1)}{(p^2-1)^2}$ and $\dfrac1p-h(p)=\dfrac{-(p^2+1)}{p(p^2-1)}$, and simplifying the quotient gives exactly \[\frac{h'(p)}{1/p-h(p)}=\frac{2p}{p^2-1}.\]

Step 6: Integrate

\texttt{sympy.integrate}$\!\left(\dfrac{2p}{p^2-1},p\right)=\ln(p^2-1)$, so $\ln|y|=\ln|p^2-1|+\text{const}$, giving \[y=C(p^2-1),\qquad x=y\,h(p)=2Cp.\]

Step 7: Gold-standard verification of the TRUE general solution

\texttt{sympy} confirms $\dfrac{dy/dp}{dx/dp}=\dfrac{2Cp}{2C}=p$ identically, and that $x(p^2-1)-2yp\equiv0$ for these formulas. Eliminating $p=x/(2C)$ gives the closed form \[y=\frac{x^2}{4C}-C,\] a genuine one-parameter family of \textbf{parabolas} — not lines.

Step 8: Proper $p$-discriminant test

Let $F(x,y,p)=x(p^2-1)-2yp$. Then $\partial F/\partial p=2xp-2y=0\Rightarrow p=y/x$. Substituting back, \texttt{sympy} simplifies $F(x,y,y/x)$ to $-\dfrac{x^2+y^2}{x}$, i.e.\ the elimination forces $x^2+y^2=0$ — only the trivial point $(x,y)=(0,0)$ (the origin itself), not a curve.

Step 9: Direct-substitution test of the source's claim $y\equiv0$

The source separately argues (from the singularities of $x=2yp/(p^2-1)$ at $p=\pm1$) that $y=0$ is "the" singular solution. Testing this directly: if $y\equiv0$ then $y'=p\equiv0$, and substituting $y=0,p=0$ into the \emph{original} relation $x(p^2-1)=2yp$ gives $x(0-1)-2(0)(0)=-x$. This is \textbf{not} identically zero for all $x$ — it vanishes only at the single point $x=0$. So $y=0$ does \emph{not} satisfy the original differential relation as a curve, and the source's boundary-value argument for it is invalid.