Find the coordinates of the focus,axis of the parabola,the equation of the directrix,and the length of the latus rectum for $x^{2}=6y$.
(N/A) The given equation is $x^{2}=6y$.
Comparing this with the standard form $x^{2}=4ay$,we get $4a=6$,which implies $a=\frac{3}{2}$.
Since the coefficient of $y$ is positive,the parabola opens upwards.
$1$. Coordinates of the focus: $(0, a) = (0, \frac{3}{2})$.
$2$. Axis of the parabola: Since the equation involves $x^{2}$,the axis is the $y$-axis $(x=0)$.
$3$. Equation of the directrix: $y = -a$,so $y = -\frac{3}{2}$.
$4$. Length of the latus rectum: $4a = 6$.
Comparing this with the standard form $x^{2}=4ay$,we get $4a=6$,which implies $a=\frac{3}{2}$.
Since the coefficient of $y$ is positive,the parabola opens upwards.
$1$. Coordinates of the focus: $(0, a) = (0, \frac{3}{2})$.
$2$. Axis of the parabola: Since the equation involves $x^{2}$,the axis is the $y$-axis $(x=0)$.
$3$. Equation of the directrix: $y = -a$,so $y = -\frac{3}{2}$.
$4$. Length of the latus rectum: $4a = 6$.
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