Q.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.

Complete the square in $y$ to write the parabola in standard form $(y-k)^2=-4a(x-h)$, then read off every required element.

1. Given equation.

$y^2+8x-6y+1=0$

2. Complete the square in $y$.

$y^2-6y=-8x-1$

$y^2-6y+9=-8x-1+9$

$(y-3)^2=-8x+8=-8(x-1)$

3. Compare with the standard form $(y-k)^2=-4a(x-h)$, which is a parabola opening in the $-x$ direction with vertex $(h,k)$:

$4a=8 \ \Rightarrow\ a=2,\qquad (h,k)=(1,3)$

4. Vertex.

$(h,k)=(1,3)$

5. Axis of the parabola (the horizontal line through the vertex, parallel to the axis of symmetry):

$y=3$

6. Focus. Since the parabola opens toward $-x$, the focus lies at $(h-a,k)$:

$(1-2,\ 3)=(-1,3)$

7. Directrix. The directrix lies at $x=h+a$:

$x=1+2=3$

8. Latus rectum. The latus rectum is the vertical line through the focus:

$x=-1$

and its length is: …