Question

Find the vertex, axis, focus, directrix, latus rectrum of following parabolas:
$x^2+2y=8x-7$

Answer 1

$x^2+2y=8x-7$
$x^2-8x=-2y-7$
$\Rightarrow x^2-2\times 4\times x+42=-2y-7+4^2$
$\Rightarrow (x-4{)}^2=-2y-7+16$
$\Rightarrow (x-4{)}^2=-2y+9$
$\Rightarrow (x-4{)}^2=-2(y-\frac{9}{2})$ ....(i)
For replacing origin $(0,0)$ at $(4,\frac{9}{2})$, put $x-4=X$ and $y-\frac{9}{2}=Y$
$X^2-2Y\Rightarrow X^2=\frac{1}{2}\times T$ ....(ii)
This is of the form $X^2=4ay$ where $a=-\frac{1}{2}$
In parabola $X^2=4ay$
Axis $X=0$, then $x-4=0\Rightarrow x=4$
Vertex $(0,0)$, then
$x-4=0\Rightarrow x=4$
and $y-\frac{9}{2}=0\Rightarrow y=\frac{9}{2}$
Thus vertex is $(4,9/2)$
Focus $(0,-a)$, then
$x-4=0\Rightarrow x=4$
and $y-\frac{9}{2}=-\frac{1}{2}\Rightarrow y=\frac{9}{2}-\frac{1}{2}=4$
Thus focus is $(4,4)$.
Latus rectum $=4a=4\times \frac{1}{2}=2$.