Through an arbitrary point M of a curve c in the plane z=0…

Through an arbitrary point \(M\) of a curve \(c\) in the plane \(z=0\), a line \(MN\) is drawn such that the ratio of the slope of \(MN\) to the slope of the tangent to \(c\) at \(M\) is a constant real number \(a\). \(1^{\circ}\) Find the equation of curve \(c\) if the line \(MN\) is always tangent to the parabola \(y=\tfrac12x^2\). Distinguish the cases \(a=\tfrac12\) and \(a\neq\tfrac12\). \(2^{\circ}\) Find the equation of the ruled surface whose director plane is the coordinate plane \(z=0\) and whose generators are lines through the \(Oz\) axis and the curve \((L):\ x=\alpha(u),\,y=\beta(u),\,z=u\), if the projection of \((L)\) onto \(z=0\) is the curve \(c\) corresponding to \(a=\tfrac12\), where \(\mu\) is the parameter appearing in the equation of \(c\).

1. Re-derive the tangency condition from scratch

Line \(MN\): \(Y=m(X-x)+y\) with \(m=ap\) (\(p=y'\) of \(c\) at \(M\)). Tangency to \(Y=X^2/2\): substituting and collecting \(\texttt{sympy.Poly}\) coefficients in \(X\) of \(\tfrac12X^2-m(X-x)-y\) gives \((\tfrac12,\,-m,\,mx-y)\), whose discriminant Sympy expands to \(m^2-2mx+2y\). With \(m=ap\) this is \[ a^2p^2-2apx+2y=0, \qquad(\dagger) \] matching the source exactly — the setup is correct.

2. Case \(a=\tfrac12\): the source's answer fails the ODE

\((\dagger)\) becomes \(p^2-4px+8y=0\), so \(p=2x\pm2\sqrt{x^2-2y}\); take the "\(+\)" branch \(dy/dx=2x+2\sqrt{x^2-2y}\). Substituting the source's claimed \(y=\tfrac12(x-\mu)^2\) (so \(dy/dx=x-\mu\), \(x^2-2y=2\mu x-\mu^2\)) gives, via Sympy, \[ \frac{dy}{dx}-\Big(2x+2\sqrt{x^2-2y}\Big) = -\mu-x-2\sqrt{2\mu x-\mu^2}\ \not\equiv\ 0, \] only vanishing (after squaring) at the two isolated points \(x=\mu,5\mu\) — not identically. So the source's "boxed" family, presented with no integration shown, does not solve the ODE it derived.

3. The correct reduction is homogeneous, not directly separable

With \(u=\sqrt{x^2-2y}\) (so \(y=(x^2-u^2)/2\), \(y'=x-uu'\)), the "\(+\)" branch becomes \(x-uu'=2x+2u\), i.e. \(u\,du/dx=-(x+2u)\) — this is the equation the source wrote as "\(u\,du/(x+2u)=-dx\)" and incorrectly called separable (\(x\) and \(u\) are mixed in the denominator, so it is not). It is, however, homogeneous of degree \(0\) in \((x,u)\): substituting \(z=u/x\) (\(u=xz\)) gives, after simplifying, \[ xz' = -\frac{(1+z)^2}{z} \qquad\Longrightarrow\qquad \frac{z\,dz}{(1+z)^2} = -\frac{dx}{x}, \] which genuinely separates. Sympy integrates the left side to \(\ln(1+z)+\dfrac{1}{1+z}\). Writing \(1+z=(x+u)/x\) and simplifying, all \(\ln x\) terms cancel, leaving the implicit general solution \[ \ln(x+u) + \frac{x}{x+u} = C. \] Sympy verifies this rigorously: implicitly differentiating \(F=\ln(x+u)+\dfrac{x}{x+u}-C=0\) with \(u=\sqrt{x^2-2y(x)}\) and solving for \(y'(x)\) returns exactly \(2x+2\sqrt{x^2-2y}\) — the original ODE, reproduced identically, with zero residual. Repeating with the "\(-\)" branch (\(dy/dx=2x-2u\)) gives, by the same method, \(\ln(u-x)-\dfrac{x}{u-x}=C\), also verified the same way.

4. An explicit parametrization for \(a=\tfrac12\), verified

Setting \(s=x+u>0\) as parameter and solving \(\ln s + x/s=C=:\mu\) for \(x\), then \(y=s(x-s/2)\), gives the explicit parametric curve \[ x(s) = s(\mu-\ln s), \qquad y(s) = s^2\Big(\mu-\tfrac12-\ln s\Big), \qquad s>0. \] Sympy confirms \(dy/dx=2s\) and \(u:=s-x\) satisfies \(u^2=x^2-2y\) identically, and hence \(dy/dx=2x+2u\) identically — a second, independent verification of the same family, now in convenient explicit form. This (correct) parametrization replaces the source's fabricated \(y=\tfrac12(x-\mu)^2\) and is what is used for Part 2 below.

5. Case \(a\neq\tfrac12\), by the same method

\((\dagger)\) gives \(p=(x\pm u)/a\), \(u=\sqrt{x^2-2y}\). The same substitution \(z=u/x\) turns the "\(+\)" branch into \(z\,dz/\big(az^2+z+(1-a)\big)=-dx/(ax)\), which Sympy integrates; after writing \(z\pm1\) in terms of \(u,x\), the \(\ln x\) terms again cancel exactly (this cancellation is what forces \(a=\tfrac12\) to be treated separately — the combination degenerates there), giving \[ a\ln(x+u) + (a-1)\ln\big((1-a)x+au\big) = C. \] Sympy's implicit differentiation confirms this reproduces \(y'=(x+u)/a\) exactly. The source's claimed \(y=\tfrac{x^2}{2}-\tfrac{\mu x}{a}+\tfrac{\mu^2}{2a^2}\) is checked the same way and found not to satisfy \(y'=(x+u)/a\) (nonzero residual), confirming it is wrong too. The "\(-\)" branch, by the identical method, gives \(a\ln(u-x)+(a-1)\ln\big(au-(1-a)x\big)=C\), also verified.

6. Part 2: setting up the conoid

A "director plane" \(z=0\) means every generator is parallel to \(z=0\), i.e. horizontal; combined with "generators pass through the \(Oz\) axis and through \((L)\)", the generator through the point \((\alpha(u),\beta(u),u)\) of \((L)\) must meet the axis at the same height \((0,0,u)\) (a line through points at two different heights could not be horizontal). So each generator, for parameter \(u=z\), is \[ (X,Y,Z) = \big(t\,\alpha(z),\ t\,\beta(z),\ z\big), \qquad t\in\mathbb R. \] Since the projection of \((L)\) onto \(z=0\) is curve \(c\) for \(a=\tfrac12\), take (using the verified Step 4 parametrization with the surface parameter renamed \(z\)): \[ \alpha(z)=z(\mu-\ln z), \qquad \beta(z)=z^2\Big(\mu-\tfrac12-\ln z\Big). \]

7. Eliminate the parameters and verify

From \(X=t\alpha(z),\,Y=t\beta(z)\): \(Y/X=\beta(z)/\alpha(z)\), independent of \(t\); together with \(Z=z\) this eliminates both parameters, giving \(Y\alpha(Z)=X\beta(Z)\), i.e. (dividing by \(Z\)) \[ \boxed{\ y\big(\mu-\ln z\big) = xz\Big(\mu-\tfrac12-\ln z\Big)\ } \] Sympy substitutes \(X=t\alpha(z),Y=t\beta(z),Z=z\) directly into \(Y(\mu-\ln Z)-XZ(\mu-\tfrac12-\ln Z)\) and simplifies the result to \(0\) identically in \(t,z,\mu\) — confirming the surface equation holds for every generator. As spot checks: points of \((L)\) itself (\(t=1\)) satisfy it trivially by construction, and every point of the \(Oz\) axis (\(x=y=0\)) satisfies it for every \(z\), as required since the axis lies on the surface.