Imaginary vector field lines

Question

If we have some function of an ellipsoid, such as $f(x,y)=100-\frac{x^2}{25}-\frac{y^2}{6}$, whose gradient vector field is $\nabla f(x,y)=(-\frac{2x}{25}, -\frac{y}{3}$), then the field lines of $\nabla f(x(t),y(t))$ can be computed by solving $\frac{dy}{dx}=\frac{25y}{6x}$ to be $y(x) = Cx^{25/6}$, which is a family of "parabolas" defined in the 1st and 4th quadrants only (for real values).

However, the gradient vector field is orthogonal to the level curves of $f(x, y)$, and therefore it must be defined in all quadrants, and, hence, the field lines must also be defined in all quadrants. But they don't seem to be.

Does anyone know how this is possible?

Answer 1

I figured this out: the solution can be written as $y=C|x^{25}|^{1/6}$, and we're done.

Answer 2

By integration, the solution of the separable differential equation is

$$6\log(|y|)=25\log(|x|)+C,$$

$$y=\pm C|x|^{25/6}.$$

So nothing is missing.