Answer
Focus = $\left(-\dfrac{1}{128}, 0\right)$;
Directrix $x=\dfrac{1}{128}$;
Axis of symmetry $y=0$
Work Step by Step
Bring the given equation to the form $y^2=4px$:
$y^2=-\frac{1}{32}x$
Here, $4p=-\dfrac{1}{32} \implies p=-\dfrac{1}{128}$ and vertex $(h,k)=(0,0)$
Since, the parabola is horizontal, the axis of symmetry is $y=0$
with focus $(p,0)=\left(-\dfrac{1}{128}, 0\right)$
Now, the directrix is $x=-p=-\left(-\dfrac{1}{128}\right)=\dfrac{1}{128}$
Our results are:
Focus = $\left(-\dfrac{1}{128}, 0\right)$;
Directrix $x=\dfrac{1}{128}$;
Axis of symmetry $y=0$.
