Find the coordinates of the focus,axis of the parabola,the equation of the directrix,and the length of the latus rectum for $x^{2} = -16y$.
(N/A) The given equation is $x^{2} = -16y$.
Here,the coefficient of $y$ is negative,so the parabola opens downwards.
Comparing this equation with the standard form $x^{2} = -4ay$,we obtain:
$-4a = -16 \Rightarrow a = 4$.
$\therefore$ The coordinates of the focus are $(0, -a) = (0, -4)$.
Since the equation involves $x^{2}$,the axis of the parabola is the $y$-axis (i.e.,$x = 0$).
The equation of the directrix is $y = a$,which is $y = 4$.
The length of the latus rectum is $4a = 4(4) = 16$.
Here,the coefficient of $y$ is negative,so the parabola opens downwards.
Comparing this equation with the standard form $x^{2} = -4ay$,we obtain:
$-4a = -16 \Rightarrow a = 4$.
$\therefore$ The coordinates of the focus are $(0, -a) = (0, -4)$.
Since the equation involves $x^{2}$,the axis of the parabola is the $y$-axis (i.e.,$x = 0$).
The equation of the directrix is $y = a$,which is $y = 4$.
The length of the latus rectum is $4a = 4(4) = 16$.
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