Help me find these? Vertex, Focus, Directrix, and Length of focal chord. for y^2+6y-2x+13=0 Any help is appreciated!

Question

campbell_st · Answer

complete the square in y after rewriting the equation as 

$$y^2 + 6y = 2x - 13 $$

campbell_st · Answer

this is a parabola which is concave right... 


the general form is 

$$(y - k)^2 = 4a(x - h)$$

the vertex is (h, k)     a = focal length

anonymous · Answer

so vertex would be -13,6 and focal length is 2? @campbell_st

campbell_st · Answer

Ummm no

its 

$$(y + 3)^2 = 2(x - 2)$$

so it looks like the focal length is 1/2   and the vertex is (2, -3)

campbell_st · Answer

your parabola looks like 

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