Find the axies, vertex, focus, equation of directrix , latus rectum , length of the latus rectum for the following parabolas and hence sketch their graphs. $x^{2}-6x-12y-3=0$.

This is the fifth part of the multi-part question Q2

Toolbox:
• Standard parabolas :
• (i)$y^2=4ax$
• In this case the axis of symmetry is $x$-axis.
• The focus is $F(a,0)$ and the equation of the directrix is $x=-a$
• (ii)$y^2=-4ax$
• In this case the focus is $F(-a,0)$ and the equation of directrix is $x=a$
• (iii)$x^2=4ay$
• In this case the axis of symmetry is $y$-axis.
• The focus is $F(0,a)$ and the equation of directrix is $y=-a$
• (iv)$x^2=-4ay$
• In this case the focus is $F(0,-a)$ and the equation of directrix is $y=a$
• Shifting of origin from (0,0) to $(h,k)$ then if the new axes form $X-Y$ coordinate system then
• $x=X+a$
• $y=Y+k$
• The new coordinates of a point $P(x,y)$ become $X=x-a$,$Y=y-k$.
Step 1:
$x^2-6x=12y+3$
$x^2-6x+9=12y+12$
$(x-3)^2=12(y+1)$
Shifting the origin to $(3,-1)$ by translation of axes.
$X=x-3,Y=y+1$
Step 2:
$XY$-axis :
$X^2=12Y$
The parabola opens upwards.
$4a=12$
$a=3$
Vertex : $(0,0)$
Focus : $(0,3)$
Equation of directrix : $Y=-3$
Equation of latus rectum : $Y=3$
Length of latus rectum : $12$
Step 3:
$xy$-axis :
$(x-3)^2=12(y+1)$
Vertex : $(3,-1)$
Focus : $(3,2)$
Equation of directrix : $y+1=-3\Rightarrow y=-4$
Equation of latus rectum : $y+1=3\Rightarrow y=2$
Length of latus rectum : $12$
edited Jun 14, 2013