Browse Questions

# Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum for the following : $y^2=12x$

$\begin {array} {1 1} (A)\;focus : (0,3), axis :y - axis , Equation \: of \: directrix : x - 3=0, length \: of \: the \: latucrectum : 12 \\ (B)\;focus : (3,0), axis : x - axis , Equation \: of \: directrix : x + 3=0, length \: of \: the \: latucrectum : 12 \\ (C)\;focus : (-3,0), axis : x - axis , Equation \: of \: directrix : x - 3=0, length \: of \: the \: latucrectum : 12 \\ (D)\;focus : (0,-3), axis : y - axis , Equation \: of \: directrix : x + 3=0, length \: of \: the \: latucrectum : 12 \end {array}$

Toolbox:
• Equation of a parabola is $y^2=4ax$
• If x is positive , then the parabola opens towards the right.
• If the equation involves $y^2$ then the axis of the parabola is x - axis.
• Equation of the directrix is x = -a or x+a=0
• Length of the lactus rectum is 4a.
Step 1 :
The given equation is $y^2=12x$
Since the coefficient of x is positive, the parabola opens towards the right.
We know the equation of parabola is $y^2 = 4ax$
Comparing both the equations we get,
$4a=12$
$\Rightarrow a=3$
$\therefore$ The coordinates of focus is (3, 0)
The axis of the parabola is x - axis.
Equation of directrix is x = -3 or x+3=0
Length of the latus rectum is $4 \times 3 = 12$
edited Jul 14, 2014