Find the curve such that the sum of the intercepts made on…

Find the curve such that the sum of the intercepts made on the coordinate axes by the tangent at any point equals the constant $a$.

Explanation of Method

At a point $(x,y)$ with slope $p=y'$, the tangent line $Y-y=p(X-x)$ meets the axes at $A\big(x-\tfrac{y}{p},0\big)$ and $B\big(0,\,y-xp\big)$. The stated condition is $\big(x-\tfrac{y}{p}\big)+\big(y-xp\big)=a$. Clearing the denominator turns this into a Clairaut equation, which is then solved by the standard general/singular decomposition.

Step 1: Set up and clear the denominator

Starting from $\big(x-\frac{y}{p}\big)+(y-xp)=a$ and multiplying by $p$, \texttt{sympy.expand} gives \[ -ap-p^2x+px+py-y=0. \] Solving this (via \texttt{sympy.solve}) explicitly for $y$ yields \[ y=\frac{p(a+px-x)}{p-1}, \] which is algebraically identical (residual $0$ under \texttt{simplify}) to the target form \[ y = xp+\frac{ap}{p-1}, \qquad p\ne 1, \] confirming the source's rearrangement.

Step 2: Identify the Clairaut equation and its general solution

This is Clairaut's equation $y=xp+f(p)$ with $f(p)=\dfrac{ap}{p-1}$, depending only on $p$. Replacing $p$ by the constant $c$ gives the general solution \[ y = cx+\frac{ac}{c-1}, \] whose trivial substitution residual (with $y'=c$ constant) is verified to be $0$.

Step 3: Envelope condition and solving for $p$

With $f(p)=\dfrac{ap}{p-1}$, $f'(p)=-\dfrac{a}{(p-1)^2}$, so the envelope condition $x=-f'(p)$ gives \[ x=\frac{a}{(p-1)^2}. \] Solving with \texttt{sympy.solve} returns \[ p = 1\pm\sqrt{\frac{a}{x}}, \] matching the source, and each root is confirmed (residual $0$) to satisfy $x=-f'(p)$.

Step 4: Candidate singular-solution branches

Substituting $p=1+\sqrt{a/x}$ into $y=xp+\dfrac{ap}{p-1}$ and simplifying gives $y=(\sqrt{x}+\sqrt{a})^2$; the branch $p=1-\sqrt{a/x}$ gives $y=(\sqrt{x}-\sqrt{a})^2$, as claimed by the source.

Step 5: Genuine verification of both branches (source left this step incomplete)

The source asserts these satisfy the original geometric condition "after simplification (algebraically intensive)" without showing the work. This was carried out explicitly: for each candidate, $y'$ was computed by direct differentiation and substituted into the full left side $\big(x-\tfrac{y}{y'}\big)+(y-xy')$, then reduced with \texttt{sympy.simplify}.

For $y=(\sqrt{x}+\sqrt{a})^2$: $y'=1+\sqrt{a/x}$, and the left side simplifies exactly to $a$ (residual $0$).

For $y=(\sqrt{x}-\sqrt{a})^2$: $y'=1-\sqrt{a/x}$, and the left side also simplifies exactly to $a$ (residual $0$).

Both branches genuinely satisfy the condition — the source's unverified claim turns out to be correct for both signs, not merely for the one it emphasized. Together the two branches are the two halves of the single curve \[ (\sqrt{x}-\sqrt{y})^2=a \quad\Longleftrightarrow\quad \sqrt{y}=\sqrt{x}\pm\sqrt{a}. \]

Result

Final answer:

General solution: \[ y = cx+\frac{ac}{c-1}. \] Singular solution (both sign branches independently verified by direct substitution into the original tangent-intercept condition): \[ y=(\sqrt{x}+\sqrt{a})^2 \quad\text{and}\quad y=(\sqrt{x}-\sqrt{a})^2, \] equivalently $(\sqrt{x}-\sqrt{y})^2=a$. Unlike the source, which left this verification as an unproven assertion, both branches are here confirmed exactly.