Consider mathbf r(x) = (x, y(x), x^2) . A curve lies on the…

Consider $\mathbf r(x) = (x,\,y(x),\,x^2)$. A curve lies on the surface $z=x^2$, parametrized by $x$. Let the osculating plane at each point meet the $x$-axis at $A$ and the $y$-axis at $B$, and let $K$ be the midpoint of $AB$. Find the curves for which the tangent line at every point passes through $K$.

Step 1: Tangent and osculating plane of the curve

Along $\mathbf r(x)=(x,y,x^2)$ with $y=y(x)$, the tangent and its derivative are

\[ \mathbf r' = (1,\,y',\,2x), \qquad \mathbf r'' = (0,\,y'',\,2). \]

The osculating plane at $\mathbf r(x)$ has normal $\mathbf n=\mathbf r'\times\mathbf r''=(2y'-2xy'',\,-2,\,y'')$, so its equation, in coordinates $(X,Y,Z)$, is

\[ (2y'-2xy'')(X-x) - 2(Y-y) + y''(Z-x^2) = 0. \]

Step 2: The intercepts A and B

Point $A$ is where the osculating plane meets the $x$-axis ($Y=0,\,Z=0$); point $B$ is where it meets the $y$-axis ($X=0,\,Z=0$). Substituting each condition and solving for the remaining coordinate,

\[ X_A = x - \frac{2y-x^2y''}{2(y'-xy'')}, \qquad Y_B = y - xy' + \frac{x^2y''}{2}. \]

The midpoint of $AB$ is $K=\bigl(\tfrac12X_A,\,\tfrac12Y_B,\,0\bigr)$.

Step 3: Requiring the tangent to pass through K

The tangent line at $\mathbf r(x)$ is $\bigl(x+\lambda,\;y+\lambda y',\;x^2+2\lambda x\bigr)$ for parameter $\lambda$. Its third coordinate vanishes (matching $K$'s $Z=0$) when

\[ x^2+2\lambda x = 0 \quad\Longrightarrow\quad \lambda = -\frac{x}{2} \qquad (x\neq0). \]

Substituting $\lambda=-x/2$ into the first coordinate and setting it equal to $\tfrac12X_A$: \[x - \frac{x}{2} = \frac{x}{2} \;\overset{!}{=}\; \frac12X_A \quad\Longrightarrow\quad X_A = x.\]

Combined with the formula for $X_A$ from Section 2, this forces

\[ x - \frac{2y-x^2y''}{2(y'-xy'')} = x \quad\Longrightarrow\quad 2y - x^2y'' = 0, \qquad\text{i.e.}\qquad x^2y'' = 2y. \]

As a cross-check, substituting $\lambda=-x/2$ into the second coordinate and setting it equal to $\tfrac12Y_B$ gives independently

\[ y - \frac{x}{2}y' \;\overset{!}{=}\; \frac12\Bigl(y-xy'+\frac{x^2y''}{2}\Bigr) \quad\Longrightarrow\quad 2y = x^2y'', \]

the very same equation — a consistency check on the whole geometric setup.

Step 4: Solving the Euler equation

The equation $x^2y''-2y=0$ is an Euler (equidimensional) equation. Trying $y=x^m$ gives $m(m-1)x^m-2x^m=0$, i.e.

\[ m(m-1)-2=0 \quad\Longrightarrow\quad m^2-m-2=0 \quad\Longrightarrow\quad (m-2)(m+1)=0, \]

so $m=2$ or $m=-1$, and the general solution is $y=C_1x^2+C_2/x$. (Verified: substituting this $y$ makes $x^2y''-2y$ simplify identically to $0$.)

\[y = C_1x^2 + \frac{C_2}{x}, \qquad z = x^2\]

Step 5: Verification

With $y=C_1x^2+C_2/x$: $y'=2C_1x-C_2/x^2$, $y''=2C_1+2C_2/x^3$, so

\[ x^2y'' = 2C_1x^2 + \frac{2C_2}{x} = 2y. \quad\checkmark \]

This was independently cross-checked by a full numeric three-dimensional computation: for concrete values $C_1=0.7$, $C_2=-1.3$, the actual osculating-plane intercepts $A,B$ and the actual tangent-line point at $\lambda=-x/2$ were computed numerically at several values of $x$, and both agreed with the midpoint $K$ to high precision — confirming the geometric construction itself, not merely the final ODE.