A downward opening parabola described by $x=2at$ and $y=-a t^2$ is perturbed by adding a small quantity, proportional to the y-coordinate, in the expression for $x$. The perturbed curve is $x=2at+at^2\delta$ and $y=-a t^2$. $0<\delta<=1$. When plotted with values $a=1$ and $\delta=0.1$, produces a perturbed parabola with its vertex close to the unperturbed parabola. See figure 1 (Blue, unperturbed. Orange, perturbed).
When converted to Cartesian coordinates, using the standard technique, a conic equation $x^2+2xy\delta +y^2+4ay=0$ is obtained (the curve is a rotated parabola). The vertex of this parabola is at $(\frac{-a}{\delta},\frac{-a}{\delta^2})$. To verify, when solved for and graphed, with previous values for $a$ and$\delta$, $y=\frac{-2a-x\delta\pm2\sqrt{a^2+ax\delta}}{\delta^2}$, a parabola with its vertex in the third quadrant is obtained as expected from the calculated vertex position for the Cartesian transformed curve. See figure 2 (Blue, unperturbed. Orange, perturbed).
Why the vertex is shifting after the coordinate transformation? Can somebody please explain where I may be committing a mistake.