Asked by Robert
                Find the vertex focus directrix and axis of symmetry of each parabola y²–8x–6y–3=0
            
            
        Answers
                    Answered by
            Reiny
            
    y²–8x–6y–3=0
looks like a horizontal parabola, let's complete the square
y^2 - 6y + 9 = 8x+3 + 9
(y - 3)^2 = 8x + 12
= 8(x + 3/2)
from that version you should be able to pick out the vertex as (-3/2 , 3)
recall that in y^2 = 4px the focus is (p,0) and the directrix is y = -p
So from you equation, 4p = 8, then p = 2
the focus is(2-3/2, 3), or (-1/2 , 3)
and the directrix is ........
https://www.wolframalpha.com/input/?i=y%C2%B2%E2%80%938x%E2%80%936y%E2%80%933%3D0
    
looks like a horizontal parabola, let's complete the square
y^2 - 6y + 9 = 8x+3 + 9
(y - 3)^2 = 8x + 12
= 8(x + 3/2)
from that version you should be able to pick out the vertex as (-3/2 , 3)
recall that in y^2 = 4px the focus is (p,0) and the directrix is y = -p
So from you equation, 4p = 8, then p = 2
the focus is(2-3/2, 3), or (-1/2 , 3)
and the directrix is ........
https://www.wolframalpha.com/input/?i=y%C2%B2%E2%80%938x%E2%80%936y%E2%80%933%3D0
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