Normals to the parabola y = x² are drawn at the points
Normals to the parabola $y = x^{2}$ are drawn at the points with abscissas $x$ and $x+h$. Find the limiting position of their intersection point as $h\to0$.
Explanation of Method
The limit is the centre of curvature, and the locus is the evolute.
Step 1: The normal at an arbitrary abscissa
At $t$ the tangent has slope $2t$, so the normal has slope $-\dfrac{1}{2t}$ and passes through $\bigl(t,t^{2}\bigr)$: \[Y = -\frac{X}{2t}+\frac12+t^{2}. \tag{1}\]
Step 2: Intersect the normals at $t = x$ and $t = x+h$
Equating the two right-hand sides of (1): $-\frac{X}{2x}+x^{2} = -\frac{X}{2\bigl(x+h\bigr)}+\bigl(x+h\bigr)^{2}$.
\[ X\left[\frac{1}{2\bigl(x+h\bigr)}-\frac{1}{2x}\right] = \bigl(x+h\bigr)^{2}-x^{2} = h\bigl(2x+h\bigr). \tag{2} \]
The bracket is $\dfrac{x-\bigl(x+h\bigr)}{2x\bigl(x+h\bigr)} = \dfrac{-h}{2x\bigl(x+h\bigr)}$ — note the minus sign, which is what puts the intersection on the far side of the axis. Dividing by $h\neq0$:
\[ X = -2x\bigl(x+h\bigr)\bigl(2x+h\bigr). \tag{3} \]
Step 3: The other coordinate
Substituting (3) into (1) with $t = x$: \[Y = \bigl(x+h\bigr)\bigl(2x+h\bigr)+\frac12+x^{2} = 2x^{2}+3xh+h^{2}+\frac12+x^{2}. \tag{4}\]
Step 4: Let $h\to0$
\[ X_{0} = -4x^{3}, \qquad Y_{0} = 3x^{2}+\frac12. \tag{5} \]
These are exactly the coordinates of the centre of curvature of the parabola at $\bigl(x,x^{2}\bigr)$: with $y' = 2x$ and $y'' = 2$, \[X_{0} = x-\frac{y'\bigl(1+y'^{2}\bigr)}{y''} = x-x\bigl(1+4x^{2}\bigr) = -4x^{3}, \qquad Y_{0} = x^{2}+\frac{1+y'^{2}}{y''} = 3x^{2}+\frac12.\]
Step 5: The locus
Eliminating $x$ from (5): $x^{2} = \dfrac{Y-\tfrac12}{3}$ and $x^{3} = -\dfrac{X}{4}$, so $x^{6}$ gives
\[ \frac{X^{2}}{16} = \left(\frac{Y-\tfrac12}{3}\right)^{3} \;\Longleftrightarrow\; 27X^{2} = 16\left(Y-\frac12\right)^{3}. \tag{6} \]
This is the evolute of the parabola, a semicubical parabola with its cusp at $\left(0,\tfrac12\right)$.
Step 6: Verification
\[x = 1:\ h = 0.1 \Rightarrow \bigl(-4.62,\,3.81\bigr); \quad h = 0.001 \Rightarrow \bigl(-4.006,\,3.503\bigr); \quad h = 10^{-6} \Rightarrow \bigl(-4.000006,\,3.500003\bigr) \;\;\checkmark\]
\[x = 2,\ h = 0.001:\ \bigl(-32.024,\,12.506\bigr) \to \bigl(-32,\,12.5\bigr) = \bigl(-4x^{3},3x^{2}+\tfrac12\bigr) \;\;\checkmark\]
\[x = 1:\ 27X^{2} = 27\cdot16 = 432 = 16\bigl(3.5-0.5\bigr)^{3} = 16\cdot27 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\text{limiting point } \left(-4x^{3},\;3x^{2}+\frac12\right)\,}\]
Final answer:
\[\boxed{\,\text{locus (the evolute): } 27X^{2} = 16\left(Y-\frac12\right)^{3}\,}\]
Remark
Remark. The sign in step 2 decides everything, and it is easy to lose. Substituting a candidate point back into a single normal proves nothing: every point of the form $\bigl(X,-\tfrac{X}{2x}+\tfrac12+x^{2}\bigr)$ satisfies that equation, including the incorrect $\bigl(4x^{3},\tfrac12-x^{2}\bigr)$. The honest checks are the numerical limit above and the independent formula for the centre of curvature, and both give $X_{0} = -4x^{3}$, $Y_{0} = 3x^{2}+\tfrac12$ — the centre lying, as it must, on the concave side of the parabola, above the curve, not below it.