what is the equation of the parabola in vertex form with focus at (2 -4) and directrix y=-6?
A.) (x+2)^2=1/4(y+4)
B.) (x-2)^2=4(y+5)
C.) (y+4)^2=-4(x-2) '
D.) (x-2)^2=-1/4(y+5)

Question

what is the equation of the parabola in vertex form with focus at (2 -4) and directrix y=-6? 
A.) (x+2)^2=1/4(y+4) 
B.) (x-2)^2=4(y+5) 
C.) (y+4)^2=-4(x-2) '
D.) (x-2)^2=-1/4(y+5)

driftracer305 · Answer

k focus (2,-4) = (h,k)

samsan9 · Answer

ok and then what?

driftracer305 · Answer

(x – h)2 = 4p(y – k) .......... so      (x-2)^2 = 4p(y+4)

samsan9 · Answer

that isnt one of the options though?

samsan9 · Answer

so A?

driftracer305 · Answer

no no........ wait  up....... something's wrong...... lemme double check

samsan9 · Answer

ok ill wait

driftracer305 · Answer

yep it's B........ it got it now.......m 5000% sure it is B

driftracer305 · Answer

@samsan9 ........sorry i messed up earlier dude......

driftracer305 · Answer

@samsan9   u there man?

samsan9 · Answer

yeah dude sorry ok ill try it

driftracer305 · Answer

no i got it......... it is B

samsan9 · Answer

hell yeah thank you dude you get a fan and a medal :D

driftracer305 · Answer

lol....... but dont u wanna knw how??  or u figured it out hmm??

samsan9 · Answer

yes i would appreciate especially with a final coming up

driftracer305 · Answer

k so focus is never (h,k) my first mistake........  nw a VERTEX is in between the focus and directrix..... here -5 between  -6 and -4.........

so p = absolute value of  -6+5  = 1

and vertex = (2,-5) = (h,k).........

(x-2)^2 = 4(1)(y+5)