A parabola whose axis is oblique to the orthogonal coordinate axes is of the form $f(x,y)= 0$, for example
$f(x,y) = 9x^2 + 24 xy + 16 y^2 + 22x + 46 y + 9=0.$
Using algebra only it is airly straightforward to find its apex once you rearrange the equation above to
$f= (3x+4y+5)^2 = 2(4x-3y+8).$
The intersection of the tangent to the summit and axis gives $\left(-\frac{47}{25},\frac{4}{25}\right)$.
I would like to get the vertex result using calculus only. I suppose the way to do is to minimize the curvature radius (unless I'm answered a better way). How do you extend the formula I know for $\{x,f(x)\}$ curves —
$R=\frac{(1+f'^2)^{3/2}}{f''}$
to this $f(x,y)=0$ curve? Thanks.