Find the locus of points such that the two integral curves…
Find the locus of points such that the two integral curves of \[ x(y')^2 - yy' + 1 = 0 \] passing through each such point intersect: \(1^{\circ}\) at right angles; \(2^{\circ}\) at an angle of \(45^{\circ}\).
Explanation of Method
The equation is quadratic in \(p=y'\); Sympy solves \(xp^2-yp+1=0\) for \(p\) directly and confirms both Vieta relations \(p_1+p_2=y/x\), \(p_1p_2=1/x\) symbolically (not just algebraically by hand). The orthogonality condition \(p_1p_2=-1\) and the \(45^\circ\) condition \(|p_1-p_2|=|1+p_1p_2|\) are then applied to the Sympy-computed roots directly, reproducing the source's loci exactly; the source's own derivation here is sound, so this entry mainly upgrades hand algebra to independent symbolic computation.
1. Solve for the two slopes and verify Vieta's formulas
Sympy's \(\texttt{solve}\) applied to \(xp^2-yp+1=0\) returns \[ p_{1,2} = \frac{y\mp\sqrt{y^2-4x}}{2x}. \] Symbolic simplification confirms \(p_1+p_2=y/x\) and \(p_1p_2=1/x\) exactly, matching Vieta's formulas for a monic-normalized quadratic \(p^2-(y/x)p+1/x=0\).
2. Orthogonal intersection (\(90^\circ\))
Perpendicularity means \(p_1p_2=-1\). Solving \(1/x=-1\) with Sympy gives the single solution \[ \boxed{\ x=-1\ } \] a vertical line, matching the source.
3. \(45^\circ\) intersection
The angle-between-lines formula requires \(|p_1-p_2|=|1+p_1p_2|\) for \(\tan45^\circ=1\). Sympy computes \(p_1-p_2=-\sqrt{y^2-4x}/x\) and \(1+p_1p_2=(x+1)/x\) directly from the Step 1 roots. Squaring the condition and simplifying \((p_1-p_2)^2-(1+p_1p_2)^2\) with Sympy gives \(\dfrac{-x^2-6x+y^2-1}{x^2}\); setting the numerator to zero reproduces \[ \boxed{\ y^2 = x^2+6x+1\ } \] equivalently \((x+3)^2-y^2=8\), a hyperbola centered at \((-3,0)\) — matching the source.
4. Domain
Both loci require \(y^2\geq 4x\) for \(p_1,p_2\) to be real (real integral curves through the point), as the source notes.
Result
Final answer:
Both of the source's final loci are confirmed correct by independent Sympy computation of the roots, Vieta relations, and both angle conditions:
\(1^{\circ}\) Orthogonal intersection: \(\boxed{\,x=-1\,}\) (a vertical line).
\(2^{\circ}\) \(45^{\circ}\) intersection: \(\boxed{\,y^2=x^2+6x+1\,}\), i.e. \(\boxed{\,(x+3)^2-y^2=8\,}\) (a hyperbola), subject to \(y^2\geq 4x\).