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

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- 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 : $12$

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