Question

Find the vertex axis, focus and length of latus rectum of the parabola
$y^2=8x+8y$

Answer 1

$y^2=8x+8y$
$\Rightarrow y^2-8y=8x$
$\Rightarrow y^2-2\times 4\times y+4^2=8x+4^2$
$\Rightarrow (y-4{)}^2=8x+16$
$\Rightarrow (y-4{)}^2=8(x+2)$ ....(i)
For replacing origin at point $(4,-2)$, put $x+2$ and $y-4=Y$
$Y^2=8W$
$\Rightarrow Y^2=\mathrm{4.2.}X$ ....(i)
Which is of the form of parabola $y^2=4ax$ where $a=2$ and axis $X=0$
Coordinates of vertex $=(0,0)$, coordinates of focus $=(0,-1)$
Length of Latus Rectum $=4\times 2=8$
for given parabola (i), put value of $X$ and $Y$ in results.
$X=0\Rightarrow x+2-0\Rightarrow x=-2$
$Y=0\Rightarrow y-4=0\Rightarrow y=4$
Thus coordinates of vertex $=(-2,4)$
Coordinates of focus
$X=0\Rightarrow x+2=2,x=0$
$Y=0\Rightarrow y-4=0,y=4$
Thus, coordinates of focus $=(0,4)$
Axis $Y=0\Rightarrow y-4=0\Rightarrow y=4$
Latus rectum $=4a=4\times 2=8$