Let the curve be $y=f\left(x\right)$

Length of subnormal is given by $y\frac{dy}{dx}$

i.e. $y\frac{dy}{dx}=x-1$

On integrating, we get

$\frac{y^2}{2}=\frac{x^2}{2}-x+C$

Given, the curve passes through $\left(1,2\right)$

i.e. $C=\frac{5}{2}$

So, $y^2=x^2-2x+5$

or $\frac{y^2}{4}-\frac{{\left(x-1\right)}^2}{4}=1$ is an equation of a hyperbola.

Its vertex would lie on $\left(1,\pm 2\right)$