Find the curves for which the sum of the normal and the…

Find the curves for which the sum of the normal and the subnormal is proportional to the abscissa: \(N+SN=kx\). For what value of the proportionality constant \(k\) is the resulting curve a genuine curve of the second degree?

1. Set up and re-derive the homogeneous ODE from scratch

With \(N=y\sqrt{1+p^2}\), \(SN=yp\) (\(p=y'\)), the condition \(N+SN=kx\) is \(y(\sqrt{1+p^2}+p)=kx\). Isolating the radical, \(\sqrt{1+p^2}=\tfrac{kx}{y}-p\), and squaring: Sympy expands \(1+p^2-\big(\tfrac{kx}{y}-p\big)^2\) and confirms the \(p^2\) terms cancel identically, leaving (after multiplying by \(y^2\)) exactly \[ y^2 - k^2x^2 + 2kxyp = 0 \quad\Longleftrightarrow\quad 2kxy\,\frac{dy}{dx} = k^2x^2-y^2, \] reproducing the source's ODE with no algebra error. Note this squaring step yields \(p\) as an explicit, single-valued function of \((x,y,k)\) — the \(p^2\) cancellation means no spurious second root of \(p\) is introduced; the only extraneous-solution risk is a possible sign mismatch between \(\sqrt{1+p^2}\) (always \(\ge 1\)) and \(\tfrac{kx}{y}-p\), addressed rigorously in Step 4.

2. Solve the homogeneous ODE two independent ways

Via \(v=y/x\): Sympy confirms \(\displaystyle\int\frac{2kv\,dv}{k^2-v^2(1+2k)} = -\frac{k}{1+2k}\ln\big|k^2-v^2(1+2k)\big|\) by differentiating the right-hand side and simplifying the difference from the integrand to \(0\) — the source's antiderivative is correct. Solving for the constant and back-substituting \(v=y/x\) and clearing the \(x^2\) factor gives the implicit general solution \[ k^2x^2-(1+2k)y^2 = A\,x^{-1/k}\qquad(\ast) \] (the exponent \(-\tfrac{1+2k}{k}+2\) simplifies via Sympy to exactly \(-1/k\) once the equation is cleared of denominators — the source never performs this final simplification, which is exactly what is needed for the degree analysis below).

Via direct numerical integration (independent check): integrating \(y'=(k^2x^2-y^2)/(2kxy)\) numerically from \(x=1,\ y=2\) with \(k=3\) and evaluating \(A=[k^2x^2-(1+2k)y^2]\,x^{1/k}\) along the trajectory gives a constant \(A\approx-19.000000\) to six decimal places at every sampled point — confirming \((\ast)\) is the correct general solution for generic \(k\), not just an artifact of the by-hand substitution.

3. Determine which \(k\) give a genuine second-degree curve

For \((\ast)\) to be an honest polynomial equation of degree \(2\) in \(x,y\) (not a transcendental or fractional-power relation), \(-1/k\) must be a positive integer, and the coefficient \(1+2k\) of \(y^2\) must be nonzero (else the curve loses its \(y\)-dependence entirely). Checking the candidates:

• \(k=2\) (the source's final answer): \(-1/k=-\tfrac12\), not an integer — \((\ast)\) is \(4x^2-5y^2=Ax^{-1/2}\), a fractional-power (non-polynomial) relation. The source's own boxed answer \(k=2\Rightarrow x^2-y^2=C\) is not even the correct family for \(k=2\) (Sympy confirms \(4x^2-5y^2\ne C\cdot x^2-y^2\) in any case) and is structurally impossible by the source's own formula — this confirms the guess was never checked.
• \(k=-\tfrac12\): here \(1+2k=0\), so \((\ast)\) degenerates to \(\tfrac14x^2 = Ax^2\), an identity with no \(y\)-dependence at all — not a genuine one-parameter family of curves, so this candidate (also unverified by the source) fails too.
• \(k=-1\): \(-1/k=1\) (a positive integer) and \(1+2k=-1\neq0\). \((\ast)\) becomes \(x^2+y^2 = Ax\), a genuine, non-degenerate family of circles through the origin, centered on the \(Ox\)-axis. This is the only simple value of \(k\) giving a bona fide second-degree curve.

4. The critical check against the ORIGINAL (unsquared) equation — and the branch subtlety it reveals

Algebraically, \(\tfrac{kx}{y}-p = \dfrac{k^2x^2+y^2}{2kxy}\) (Sympy simplification of the explicit \(p\) from Step 1), so its sign is exactly \(\operatorname{sign}(kxy)\). Squaring is only reversible where this quantity is \(\ge 1>0\); where \(kxy<0\) instead, the curve solves the sign-flipped relation \(y(p-\sqrt{1+p^2})=kx\), i.e. \(N-SN=kx\), not the required sum equation.

For \(k=-1\), parametrizing the full circle family \(x^2+y^2=Ax\) as \(y=\sqrt{Ax-x^2}\) and computing \(p=y'\) symbolically in Sympy:

• Left half of the family (\(A<0\), i.e. \(x\le 0\), substituting \(A=-B,\,x=-t\) with \(B,t>0\)): Sympy simplifies \(y(\sqrt{1+p^2}+p)\) exactly to \(t=-x=kx\) (with \(k=-1\)) — an exact symbolic identity, confirmed for every point of this half of the family, not just a numeric spot check.
• Right half (\(A>0\), \(x\ge0\)): the same symbolic computation instead gives \(y(\sqrt{1+p^2}+p) = -x+A\ne kx\) in general; numerically, \(\tfrac{kx}{y}-p\) is verified negative at every sampled point of this half, so it solves \(N-SN=x\) (i.e. the reflected equation with the opposite sign convention), not the stated sum equation.

So the literal sum equation \(N+SN=kx\) is satisfied exactly, end-to-end, symbolically, by \(k=-1\) on the branch \(x\le0\) of the circle family \(x^2+y^2=Ax\) — the same geometric family of circles through the origin (centered on the \(Ox\)-axis) that the exponent analysis in Step 3 singled out as the only non-degenerate second-degree candidate. The mirror branch \(x\ge0\) of the identical curve family instead satisfies the companion identity \(N-SN=x\), which is the well-known reason such problems are conventionally stated up to reflection/sign of \(k\) — but it does not change the answer to "for what \(k\) is the curve second-degree": that value is \(k=-1\), verified against the original equation, not \(k=2\).