In the plane, a point O and a line d are given. 1^(°)…

In the plane, a point \(O\) and a line \(d\) are given. \(1^{\circ}\) Determine the curves \(c\) such that the distance from each point \(M\) of the curve to the line \(d\) equals the distance from \(O\) to the tangent to the curve at \(M\). \(2^{\circ}\) Through an arbitrary point \(A\) of the plane there generally pass two integral curves of the resulting differential equation. Find the locus of points \(A\) where the tangents of the two integral curves coincide.

1. Re-derive the squared-distance relation and the quadratic in \(p\) from scratch

With \(O=(0,0)\), \(d:x=a\), \(M=(x,y)\): \(\mathrm{dist}(M,d)=|x-a|\), and the tangent line \(pX-Y+(y-xp)=0\) is at distance \(|y-xp|/\sqrt{1+p^2}\) from \(O\). Squaring the condition \(\mathrm{dist}(M,d)=\mathrm{dist}(O,\text{tangent})\) gives \((a-x)^2(1+p^2)=(y-xp)^2\). Expanding this with Sympy's \(\texttt{Poly}\) and collecting powers of \(p\) reproduces, independently, exactly the source's quadratic \[ a(a-2x)\,p^2 + 2xy\,p + \big[(a-x)^2-y^2\big] = 0. \qquad (\star) \] So the setup through this point is correct; the error is entirely in what happens next.

2. The source's guessed answer fails its own equation

The source asserts, without solving \((\star)\), that the curves are parabolas \(y^2=2a(a-x)\) (inconsistently also written \(y^2=4p(a-x)\) with \(p\) reused as both slope and a free parameter). Differentiating \(y^2=2a(a-x)\) gives \(p=-a/y\); substituting \(x,y,p\) into the left side of \((\star)\) with Sympy gives \[ a(a-2x)p^2+2xyp+(a-x)^2-y^2 \;=\; \frac{a^3/2+2a^2x-3ax^2+x^3}{x-a}\ \neq\ 0, \] not identically zero. So the guessed family does not even solve the source's own correctly-derived quadratic — confirming it must be discarded, not patched.

3. The correct method: tangential (Lagrange) coordinates

Write the tangent line at parameter \(p\) as \(Y=pX+c(p)\); the point of tangency is then \(x(p)=-c'(p)\), \(y(p)=c(p)-p\,c'(p)\) (from \(Y=pX+c(p)\) together with \(dy/dp=p\,dx/dp\), which forces \(0=x+c'(p)\) — verified directly by differentiating). Then \(\mathrm{dist}(M,d)=|a-x(p)|=|a+c'(p)|\) and \(\mathrm{dist}(O,\text{tangent})=|c(p)|/\sqrt{1+p^2}\), so the geometric condition becomes, on the branch \(c(p)>0\), \[ c'(p) - \frac{c(p)}{\sqrt{1+p^2}} = -a, \] a genuine first-order linear ODE in \(c(p)\) — unlike \((\star)\), this is directly solvable. Sympy's \(\texttt{dsolve}\) (integrating factor \(\sqrt{1+p^2}-p=e^{-\operatorname{arcsinh}p}\)) returns \[ c(p) = C_1 e^{\operatorname{arcsinh}p} - \tfrac{a}{2}\big[p\operatorname{arcsinh}p+p+\sqrt{1+p^2}\operatorname{arcsinh}p\big]. \]

4. Simplify with \(p=\sinh t\) and verify the closed form

Substituting \(p=\sinh t\) (so \(\sqrt{1+p^2}=\cosh t\), \(\operatorname{arcsinh}p=t\)) and relabelling the constant as \(C_1=\tfrac{a}{2}\mu\), Sympy simplifies \(x(p)=-c'(p)\) and \(y(p)=c(p)-p\,c'(p)\) (after rewriting \(\sinh,\cosh,\tanh\) in exponentials and expanding) to the clean closed form \[ x(t) = \frac{a}{2}\Big[(t-\mu+2)+(t-\mu+1)\tanh t\Big],\qquad y(t) = \frac{a}{2}\Big[e^{t}-(t-\mu+1)\operatorname{sech}t\Big],\qquad p=y'=\sinh t. \] Substituting these back into \((a-x)^2(1+p^2)-(y-xp)^2\) and into \(\dot y - p\,\dot x\), Sympy reduces both, after rewriting everything in \(e^{t}\) and expanding, to exactly \(0\) — a full, independent confirmation that this parametric family genuinely solves the original geometric condition and the differential equation. \(\mu\) is the arbitrary constant of the family; \(t\) runs along each individual curve.

5. The mirror branch, and why two curves pass through each point

The governing relation \((a-x)^2(1+p^2)=(y-xp)^2\) is invariant under \((y,p)\mapsto(-y,-p)\), so reflecting the family of Step 4 across the \(x\)-axis gives a second, equally valid family (the \(c(p)<0\) branch). Together the two branches are exactly the two roots \(p_1,p_2\) of the quadratic \((\star)\) through a generic point \(A\) referenced in Part \(2^\circ\).

6. Part 2: the discriminant, corrected

Sympy computes the discriminant of \((\star)\) as a polynomial in \(x,y,a\) and factors it: \[ \Delta = (2xy)^2-4a(a-2x)\big[(a-x)^2-y^2\big] = 4(a-x)^2\big(y^2+2ax-a^2\big). \] This factorization is the key step the source skipped (it jumped straight from the unfactored \(\Delta=0\) line to a boxed \(y^2=a(a-x)\) with no algebra shown). Discarding the degenerate factor \(a-x=0\) (which is just the line \(d\) itself, not an envelope of the family), the genuine locus of coinciding tangents is \[ \boxed{\,y^2=a(a-2x)\,}, \] which differs from the source's unjustified \(y^2=a(a-x)\).

7. Independent cross-check: computing the envelope directly from the Part-1 family

As a second, fully independent check, the true envelope of the verified two-parameter family \(x(t,\mu),y(t,\mu)\) of Step 4 is where the Jacobian \(x_\mu y_t-y_\mu x_t\) vanishes. Sympy computes this Jacobian (after rewriting in exponentials) as \(\propto \mu-t-\tfrac12 e^{2t}-\tfrac32\); setting this factor to zero and substituting \(\mu=t+\tfrac12e^{2t}+\tfrac32\) back into \(x(t,\mu),y(t,\mu)\) gives, after simplification, \[ x=\frac{a(1-e^{2t})}{2},\qquad y=ae^{t}, \] and Sympy confirms \(y^2-a(a-2x)=a^2e^{2t}-a\big(a-a(1-e^{2t})\big)=0\) identically in \(t\). This is an independent derivation of the same locus \(y^2=a(a-2x)\) found in Step 6 by the discriminant method — the two completely different computations agree exactly.

8. Geometric identification

Writing \(y^2=a(a-2x)\) as \(y^2=-2a\big(x-\tfrac a2\big)\) exhibits it as a parabola with vertex \(\big(\tfrac a2,0\big)\), focus at \(x=\tfrac a2-\tfrac a2=0\) — i.e. exactly at \(O\) — and directrix \(x=\tfrac a2+\tfrac a2=a\) — exactly the line \(d\). So the envelope in Part 2 is precisely the parabola with focus \(O\) and directrix \(d\): the very curve the source mistakenly proposed as the answer to Part 1 turns out, correctly, to be the answer to Part 2.