The length of the subnormal is given by:
\[
\text{subnormal} = y \cdot \frac{dy}{dx} = x - 1
\]
Thus, the differential equation becomes:
\[
y \frac{dy}{dx} = x - 1 \Rightarrow y \, dy = (x - 1) \, dx
\]
Integrate both sides:
\[
\frac{y^2}{2} = \frac{(x - 1)^2}{2} + C
\Rightarrow y^2 = (x - 1)^2 + C'
\]
Use point \( (1, 2) \) to find \( C' \):
\[
4 = 0 + C' \Rightarrow C' = 4
\]
So the equation is:
\[
y^2 = (x - 1)^2 + 4 \Rightarrow (x - 1)^2 - y^2 = -4
\]
This is a conic (hyperbola). Put in standard form and find vertex:
Vertex lies at the center, \( (0, \sqrt{5}) \).