# Find the axies, vertex, focus, equation of directrix , latus rectum , length of the latus rectum for the following parabolas and hence sketch their graphs. $(x-4)^{2}=4(y+2).$

This is the third 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 $X=x-a$,$Y=y-k$.
• 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.
edited Jun 16, 2013