Question 1092269
{{{y^2-8y+12x-8=0}}}
{{{y^2-8y=-12x+8}}}
"16" will complete-the-square.
{{{y^2-8y+16=-12x+8+16}}}
{{{(y-4)^2=-12x+24}}}
{{{highlight_green(-12(x-2)=(y-4)^2)}}}


VERTEX (2,4)
FOCUS  (-1,4)
DIRECTRIX  x=5
EndPoints of LATUS RECTUM  (-1,-2) and (-1,10)





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-12(-1-2)=(y-4)^2
36=(y-4)^2
y-4=0+- 6
y=4+- 6
y=-2 or y=10