I forgot my answer:
(x+2)^2 = 4(y-5)
What is an equation of a parabola with the given vertex and focus?
vertex(-2,5)
focus(-2,6)
3 answers
If you make a sketch, you will see that the directrix must be y = 4
let P(x,y) be any point
√( (y-6)^2 + (x+2)^2 ) = √((y-4)^2 + 0)
square and expand
y^2 - 12y + 36 + (x+2)^2 = y^2 - 8y + 16
(x+2)^2 = 4y - 20
(x+2)^2 = 4(y- 5)
I agree
let P(x,y) be any point
√( (y-6)^2 + (x+2)^2 ) = √((y-4)^2 + 0)
square and expand
y^2 - 12y + 36 + (x+2)^2 = y^2 - 8y + 16
(x+2)^2 = 4y - 20
(x+2)^2 = 4(y- 5)
I agree
Thank you