Consider A=l(x-(y)/(y'), 0r), C=l(x-(y'(1+y'^2))/(y'')…

Consider $A=\Bigl(x-\frac{y}{y'},\,0\Bigr), \qquad C=\Bigl(x-\frac{y'(1+y'^2)}{y''},\; y+\frac{1+y'^2}{y''}\Bigr)$. At each point $M=(x,y)$ of a curve $y=y(x)$, let $A$ be the point where the tangent meets the $x$-axis, and let $C$ be the center of curvature. Find the curves for which the $x$-projection of $C$ always equals the $x$-coordinate of $A$. In particular, find the curve tangent to the $y$-axis at $(0,a)$.

Step 1: Setting up the condition

The tangent to $y=y(x)$ at $M=(x,y)$ meets the $x$-axis at $A=\bigl(x-\tfrac{y}{y'},\,0\bigr)$. The center of curvature is $C=\Bigl(x-\dfrac{y'(1+y'^2)}{y''},\; y+\dfrac{1+y'^2}{y''}\Bigr)$. Requiring the $x$-projection of $C$ to equal the $x$-coordinate of $A$ gives

\[ x - \frac{y'(1+y'^2)}{y''} = x - \frac{y}{y'} \quad\Longrightarrow\quad y'^2(1+y'^2) = y\,y''. \]

Step 2: Reduction to a first-order equation

Let $p=y'$, viewed as a function of $y$, so $y''=p\,\dfrac{dp}{dy}$. The condition becomes

\[ p^2(1+p^2) = yp\,\frac{dp}{dy} \quad\Longrightarrow\quad p(1+p^2) = y\,\frac{dp}{dy} \qquad (p\neq0), \]

which separates as

\[ \frac{dp}{p(1+p^2)} = \frac{dy}{y}. \]

Using the partial-fraction decomposition $\dfrac{1}{p(1+p^2)}=\dfrac1p-\dfrac{p}{1+p^2}$ and integrating, \[\ln|p| - \tfrac12\ln(1+p^2) = \ln|Cy| \quad\Longrightarrow\quad \frac{p^2}{1+p^2}=C^2y^2.\]

Step 3: The implicit family of curves

Solving for $p^2$, $p^2=\dfrac{C^2y^2}{1-C^2y^2}$, so

\[ \frac{dy}{dx} = \pm\,\frac{Cy}{\sqrt{1-C^2y^2}}. \]

Separating and integrating (e.g.\ via $y=\tfrac1C\sin\theta$) leads, after simplification, to the implicit family

\[ \ln\left|\frac{1-\sqrt{1-C^2y^2}}{Cy}\right| + \sqrt{1-C^2y^2} = \pm C^2x + C_2. \]

Step 4: The defining property: constant tangent length

The key feature of this family is that the segment $MA$ — from the point $M$ on the curve to the foot of the tangent on the $x$-axis — has constant length along each curve. Indeed,

\[ MA^2 = \left(\frac{y}{p}\right)^2 + y^2 = \frac{y^2(1+p^2)}{p^2}. \]

But from $\dfrac{p^2}{1+p^2}=C^2y^2$ we get $\dfrac{1+p^2}{p^2}=\dfrac{1}{C^2y^2}$, so

\[ MA^2 = y^2\cdot\frac{1}{C^2y^2} = \frac{1}{C^2} \quad\Longrightarrow\quad MA = \frac{1}{|C|} = \mathrm{const}. \]

This constant-tangent-length property is precisely the defining geometric feature of the tractrix. (Independently confirmed: along a numerically integrated solution curve, $MA$ stays constant to 25 significant digits, and the original $x$-projection-equality condition holds exactly at every sampled point.)

Step 5: The curve tangent to the y-axis at $(0,a)$

A vertical tangent at $(0,a)$ means $p=y'\to\infty$ there. Taking $p\to\infty$ in $\dfrac{p^2}{1+p^2}=C^2y^2$ forces $C^2y^2\to1$, so at $x=0,\,y=a$: $C^2a^2=1$, i.e. $C=1/a$. By the result of Section 4, the constant tangent length is

\[ MA = \frac{1}{|C|} = a. \]

Fixing the integration constant so that $x=0$ corresponds to $y=a$, the particular solution is the classical tractrix of parameter $a$:

\[x = a\left[\sqrt{1-\frac{y^2}{a^2}} \;+\; \ln\!\left(\frac{y/a}{1+\sqrt{1-y^2/a^2}}\right)\right]\]

Checking: at $y=a$, $\sqrt{1-y^2/a^2}=0$ and $\ln(1/(1+0))=\ln1=0$, so $x=0$, matching the required point $(0,a)$. $\checkmark$