SOLUTION: Can I have some help with this one too please? What is the equation of the parabola, with vertex at (2,-4) and directrix y=-6 A.(x+2)^2=8(y+4) B.(y+4)^2=8(x-2) C.(y+6)^2=-8(x

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Can I have some help with this one too please? What is the equation of the parabola, with vertex at (2,-4) and directrix y=-6 A.(x+2)^2=8(y+4) B.(y+4)^2=8(x-2) C.(y+6)^2=-8(x      Log On


   

Question 1100547: Can I have some help with this one too please? What is the equation of the parabola, with vertex at (2,-4) and directrix y=-6
A.(x+2)^2=8(y+4)
B.(y+4)^2=8(x-2)
C.(y+6)^2=-8(x+2)
D.(x-2)^2=8(y+4)
Found 2 solutions by richwmiller, josgarithmetic:
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
vertex at (2,-4)
(x-h)^2=4p(y-k)
(x-2)^2=4p(y+4)
Without even calculating p the only option that matches is D.(x-2)^2=8(y+4)
and yes p=2

Answer by josgarithmetic(39621) About Me  (Show Source):
You can put this solution on YOUR website!
Vertex given means you want a factor of x-2 and y+4. Only choice D has these.

Otherwise you can find the focus and use definition of parabola and given directrix and found focus and derive the equation using definition of parabola and distance formula.