The equation below describes a conic with oblique axis:
$$9 + 22 x + 9 x^2 + 46 y + 24 x y + 16 y^2=0$$
It is a parabola, as the coefficients in $x^2$, $y^2$, $xy$ form a perfect square. To find the coordinates of its vertex, it's either the hard way using algebra only on paper (I summarise that at the end of my entry) or possibly an easy way with Mathematica.
Let me describe my Mathematica approach:
Looking at the $x^2$ and $y^2$ coefficients I know that the slope of the tangent to the parabola at its vertex is $4/3$ so this tangent is of the form $y = 4x/3 + b$ and it intercepts the parabola at its vertex $(x,y)$ exactly one time. When I translate this in Mathematica terms, I have the following system of equations to solve:
{x, y, b} /.
Solve[{9 + 22 x + 9 x^2 + 46 y + 24 x y + 16 y^2 == 0,
y - (4/3)x - b == 0}, {x, y, b}]
which is not enough of course to find unique values for $(x,y,b)$.
Is it possible to add conditions to the above Solve
expressions such as only one solution for $(x,y,b)$ is to be returned (or more exactly as there are squares in the expression) two solutions but identical?.
On paper, I transformed with factors the equation of the parabola. End result put into Mathematica:
eq1 = (3 x + 4 y + 5)^2 - 2 (4 x - 3 y + 8)
eqc = 9 + 22 x + 9 x^2 + 46 y + 24 x y + 16 y^2
eqc == eq1 // FullSimplify
tangentvertex = Reduce[4 x - 3 y + 8 == 0, y]
vertex = {x, y}/.Solve[{eqc == 0, y - 8/3 - (4 x)/3 == 0}, {x, y}] // FullSimplify
Vertex coordinates at last:
{-47/25, 4/25}
Thanks for any answer.