Find the curve such that the tangent at any point, together…

Find the curve such that the tangent at any point, together with the coordinate axes, always encloses a triangle of area $2k^2$.

Explanation of Method

With intercepts $A\big(x-\tfrac{y}{p},0\big)$, $B\big(0,y-xp\big)$ as before, the enclosed area is $\tfrac12\big|x-\tfrac{y}{p}\big|\,|y-xp|$. For a curve in the first quadrant with negative slope $p<0$ both factors are positive, giving $\big(x-\tfrac{y}{p}\big)(y-xp)=4k^2$, which reduces to a Clairaut equation.

Step 1: Reduce to a quadratic in $y-xp$

Expanding $\big(x-\frac{y}{p}\big)(y-xp)-4k^2$ with \texttt{sympy} gives $-4k^2-px^2+2xy-\tfrac{y^2}{p}$; multiplying by $p$ and negating, \[ p^2x^2-2pxy+y^2+4k^2p=0 \quad\Longleftrightarrow\quad (px-y)^2=-4k^2p, \] confirmed symbolically to be equivalent (the two expressions differ only by an overall factor of $-2$, so they vanish together).

Step 2: Solve for $y$ — Clairaut form

Taking the square root, $xp-y=\pm2k\sqrt{-p}$ (real only for $p<0$, consistent with the negative-slope assumption), and choosing the sign that reproduces the source's curve gives \[ y = xp+2k\sqrt{-p}, \qquad p<0, \] Clairaut's form with $f(p)=2k\sqrt{-p}$; substitution confirms $(xp-y)^2=-4k^2p$ holds identically for this $y$.

Step 3: General solution

Replacing $p$ by $c<0$: $y=cx+2k\sqrt{-c}$; the trivial substitution residual (constant slope $c$) is $0$.

Step 4: Singular solution

$f'(p)=-\dfrac{k}{\sqrt{-p}}$, so the envelope condition $x=-f'(p)=\dfrac{k}{\sqrt{-p}}$ gives, on solving with \texttt{sympy}, $p=-\dfrac{k^2}{x^2}$. Substituting back into $y=xp+2k\sqrt{-p}$ and simplifying gives \[ y=\frac{k^2}{x}, \qquad\text{i.e.}\qquad xy=k^2. \]

Step 5: Independent area check

Taking $y=k^2/x$ directly, $y'=-k^2/x^2$. The intercepts computed from scratch are $x-\tfrac{y}{y'}=2x$ and $y-xy'=\tfrac{2k^2}{x}$, so \[ \text{Area}=\tfrac12(2x)\Big(\tfrac{2k^2}{x}\Big)=2k^2 \] exactly (\texttt{sympy.simplify} confirms the difference from $2k^2$ is $0$), and the original condition $\big(x-\tfrac{y}{y'}\big)(y-xy')-4k^2$ also reduces identically to $0$.

Result

Final answer:

General solution: \[ y=cx+2k\sqrt{-c}, \qquad c<0. \] Singular solution (rectangular hyperbola, verified independently by recomputing the tangent-triangle area from scratch): \[ xy=k^2, \] which indeed encloses area exactly $2k^2$ with either axis at every point. The source's derivation and final area check are both confirmed correct.