This is the third 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 $X=x-a$,$Y=y-k$.
- General forms of the parabola when the origin is shifted to $(h,k)$ and the new axes are parallel to the $xoy$ axes.
- Equations with respect to $xoy$ axes :
- $(y-k)^2=4a(x-h)$
- $(y-k)^2=-4a(x-h)$
- $(x-h)^2=4a(y-k)$
- $(x-h)^2=-4a(y-k)$

Step 1:

Shifting the origin to $(4,-2)$ by translation of axes $X=x-4$,$Y=y+2$.

$XY$ axis :

$X^2=4Y$

$xy$ axis :

$(x-4)^2=4(y+2)$

The parabola opens upwards.

Step 2:

$XY$ axis :

$4a=4$

$a=1$

Axis : $X=0$

Vertex : $(0,0)$

Focus : $(0,1)$

Equation of directrix : $Y=-1$

Equation of LR : $Y=1$

LR=4.

Step 3:

$xy$ axis :

Axis : $x-4=0$

Vertex : $(4,-2)$

Focus : $(4,-4)$

Equation of directrix : $y+2=-1\Rightarrow y=-3$

Equation of LR : $y+2=1\Rightarrow y=-1$

LR=4.

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