How to find the vertex of a parabola given by the bivariate quadratic $(ax+my+n)^2-4ay=0$

Question

I tried the usual rotation of the axis technique; but it rotates the curve at the coordinate origin producing wrong vertex coordinates. It appears that the parabola need be first translated to the origin and for that I need the coordinates of the vertex. An algebraic solution is preferred. Thanks and any help is appreciated.

Answer 1

The equation of the parabola is already in a convenient form for doing this. The squared linear term gives you the direction of the parabola’s axis: it is parallel to the line $ax+my=0$. The vertex is the point on the parabola at which the tangent is perpendicular to its axis. Equivalently, it’s the point at which the normal to the parabola is parallel to the axis. This condition can be expressed as $(a,m)\cdot\nabla f(x,y)=0$, where $f(x,y)$ is the left-hand side of the parabola’s equation. In fact, this is an equation of the parabola’s axis. Compute the intersection of this line with the parabola.

If you want a purely algebraic solution, you can use the fact that the tangent to the parabola at a point is the point’s polar line, then use the perpendicularity condition to generate essentially the same equation as above.