Given the function $f(x,y) = 9x^2-24xy+16y^2-20x-15y-250 \:$ find the expression for the contour corresponding to $f(x,y) = 0$.
My attempt
Using Maple to solve the equation $f(x,y) = 0$ with repsect to x and y yields the following.
$$(x,y)=\Bigg(\frac{4y}{3}+\frac{10}{9}+\frac{5\sqrt{15y+94}}{9}, y \Bigg)$$ and $$(x,y)=\Bigg(\frac{4y}{3}+\frac{10}{9}-\frac{5\sqrt{15y+94}}{9}, y \Bigg)$$
Plotting these together gives a parabola: -
I need to find the vertex and symmetry axis of this parabola, however neither of those are particularly obvious because the parabola is slanted.
My question is, how can I find the expression for this parabola?
For a rotated parabola in the form
$(ax+by+c)^2 = Ax + By +C$ such that $ax + by + c = 0$ and $Ax + By + C = 0$ are perpendicular to each other, $ax + by + c = 0$ is the line of symmetry and $Ax + By + C = 0$ is the tangent at the vertex.
Now the given parabola is $(3x-4y)^2 = 20 x + 15y + 250$. Notice that $3x - 4y = 0$ and $20x + 15y+250=0$ are already perpendicular to each other.
So the axis of symmetry is $3x-4y = 0$ and the tangent at the vertex is $4x + 3y + 50 = 0$. Intersection of both lines will give you the coordinates of the vertex.
Otherwise you would write the parabola as
$(3x-4y + c)^2 = (20 + 6c) x + (15 - 8c) y + 250 + c^2$ and solve for $c$ using $m_1 m_2 = - 1$ where $m_1$ and $m_2$ are slopes of lines $3x - 4y + c = 0$ and $(20+6c) x + (15 - 8c) y + 250 + c^2 = 0$