Find the coordinates of the focus,axis of the parabola,the equation of the directrix,and the length of the latus rectum for $y^{2}=12x$.
(N/A) The given equation is $y^{2}=12x$.
Here,the coefficient of $x$ is positive. Hence,the parabola opens towards the right.
On comparing this equation with $y^{2}=4ax$,we obtain:
$4a=12 \Rightarrow a=3$.
$\therefore$ Coordinates of the focus $= (a, 0) = (3, 0)$.
Since the given equation involves $y^{2}$,the axis of the parabola is the $x$-axis.
Equation of the directrix is $x = -a$,i.e.,$x = -3$ or $x + 3 = 0$.
Length of the latus rectum $= 4a = 4 \times 3 = 12$.
Here,the coefficient of $x$ is positive. Hence,the parabola opens towards the right.
On comparing this equation with $y^{2}=4ax$,we obtain:
$4a=12 \Rightarrow a=3$.
$\therefore$ Coordinates of the focus $= (a, 0) = (3, 0)$.
Since the given equation involves $y^{2}$,the axis of the parabola is the $x$-axis.
Equation of the directrix is $x = -a$,i.e.,$x = -3$ or $x + 3 = 0$.
Length of the latus rectum $= 4a = 4 \times 3 = 12$.
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