Find the family of curves such that the tangent at an…

Find the family of curves such that the tangent at an arbitrary point, the radius vector to that point, and the \(Ox\) axis form a triangle of constant area \(a\).

1. Triangle area as a function of \(x,y,y'\)

With \(O(0,0)\), \(P(x,y)\), \(T\big(x-\tfrac{y}{y'},0\big)\), the shoelace formula gives the signed area \(\tfrac12\big[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\big]\). Substituting the three vertices and simplifying with Sympy (\(\texttt{simplify}\) on the fully substituted expression) gives exactly \[ \text{Area}_{\text{signed}} = \frac{y(y-xy')}{2y'} = \frac{y^2}{2y'}-\frac{xy}{2}, \] confirming the source's formula symbolically, with no sign or algebra slip.

2. The differential equation

Setting \(\dfrac{y^2}{2y'}-\dfrac{xy}{2}=a\) and solving for \(y'\) with Sympy's \(\texttt{solve}\) gives \(y' = \dfrac{y^2}{xy+2a}\), i.e. \((xy+2a)\,dy = y^2\,dx\), matching the source exactly. Dividing by \(y^2\) and rearranging as an equation for \(x(y)\) gives the linear ODE \[ \frac{dx}{dy}-\frac{x}{y} = \frac{2a}{y^2}. \]

3. Solve with an integrating factor

\(\mu=e^{-\int dy/y}=1/y\) turns the left side into \(\dfrac{d}{dy}\!\big(x/y\big)\); integrating \(\dfrac{2a}{y^3}\) and solving for \(x\) gives \(x=Cy-\dfrac{a}{y}\). Independently, calling Sympy's \(\texttt{dsolve}\) directly on \(\dfrac{dx}{dy}-\dfrac{x}{y}=\dfrac{2a}{y^2}\) returns \(x(y)=C_1y-\dfrac{a}{y}\) — the identical family, confirming the by-hand integrating-factor solution.

4. Verify the area really equals \(a\) on the solution

For \(x=Cy-\tfrac{a}{y}\), Sympy computes \(\dfrac{dx}{dy}=C+\dfrac{a}{y^2}\), hence \(y'=\dfrac{dy}{dx}=\dfrac{y^2}{Cy^2+a}\), and the tangent's \(x\)-intercept simplifies to \(x-\dfrac{y}{y'}=-\dfrac{2a}{y}\) exactly (a clean closed form, symbolically verified). Substituting into the signed-area formula from Step 1 and taking \(y>0\), Sympy simplifies the result to exactly \(a\) (numerically spot-checked as well, e.g. \(a=2,\ C=3,\ y=1.5\) gives area \(=2.000\)), confirming the family genuinely has the required constant-area property.