Question

write the equation for the parabola whose vertex is at (3,5) and whose directix is the line y=3.

Answer 1

hello is that the correct prob? is the directrix=y=3?

Answer 2

that is the way its written. I have choices
a. (y+5)^2=4(x-3)
b. (x-3)^2=-4(y+5)
c. x^2=-8(y+5)
d. (x-3)^2=8(y-5)
e. none of the above

Answer 3

using the formula
4p(y-k)=(x-h)^2 and vertex =V(3,5)
4p(y-5)=(x-3)^2
directrix d=3=5-p
p=5-3=2 therefore
4p(y-k)=(x-h)^2
8(y-5)=(x-3)^2 ans.....or
8y-40=x^2 -6x+9
x^2 -6x-8y+49=0 ans....

Answer 4

the parabola opens upward