Question

Find the axis, vertex, focus, equation of directrix, latus rectum, length of latus rectum for the parabola $y^2+8x-6y+1=0$ and also draw the diagram

Answer 1

$y^2-6y=-8x-1$
$y^2-6y+9=-8x+1+9$ (Adding $9$ on both sides)
$(y-3{)}^2=-8x+8$
$(y-3{)}^2=-8(x-1)$
This is of the form $Y^2=-8X$ where $X=x-1$ and $Y=y-3$
$4a=8\Rightarrow a=\frac{8}{4}=2$ This type is open leftward

	Referred to $X,Y$ axes	Referred to $x,y$ axes where $X=x-1$ and $Y=y-3$
Axis	$Y=0$	$Y=0\Rightarrow y-3=0$ $\Rightarrow y=3$
Vertex	$(0,0)$	$X=0\Rightarrow x-1=0$ $\Rightarrow x=1$ $Y=0\Rightarrow y-3=0$ $\Rightarrow y=3$ $\therefore$ Vertex is $V(1,3)$
Focus	$(-a,0)$ i.e., $(-2,0)$	$X=0\Rightarrow x-1=0$ $\Rightarrow x=1$ $Y=0\Rightarrow y-3=0$ $\Rightarrow y=3$ $\therefore$ Focus is $F(-1,3)$
Equation of Directrix	$X=a$ i.e., $X=2$	$X=2\Rightarrow x-1=2$ $\Rightarrow x=2+1$ $\Rightarrow x=3$
Equation of Latus Rectum	$X=-a$ i.e., $X=-2$	$X=-2\Rightarrow x-1=-2$ $\Rightarrow x=-2+1$ $\Rightarrow x=-1$
Length of the Latus Rectum	$4a=4\left(2\right)=8$	$8$

$\therefore$ the graph of the parabola looks as following.