Is this parametric curve a rotated parabola?

Question

I have a parametric curve that is quadratic in both x(a) and y(a). Specifically,

$$x(a) = -a^2 +150a$$ $$y(a) = -3a^2 + 500a$$ It looks like a rotated parabola. How can I be sure?

See my Maple worksheet.

Thanks, Matt

Answer 1

Note that for large $a$ the functions are asymptotic to $(x,y)=(-a^2,-3a^2)$; hence the line $y=3x$ must be the axis of symmetry if this is to be a parabola. By comparison, the standard parabola has $x=0$ as axis of symmetry. This suggests the change of coordinates $u=y-3x,\, v=x+3y$, chosen such that $u=0$ is the line $y=3x$ and the $uv$-axes are orthogonal. In these coordinates, we immediately have $$u=y-3x=50a\implies a=u/50\implies v=1650a-10a^2=31u-\frac{1}{50}u^2$$ which is clearly a parabola.

Answer 2

HINT:

Solve for $a^2,a$ in terms of $x,y$

Use $a^2=(a)^2$ to eliminate $a$

Finally use Rotation of axes to eliminate $xy$ term