SOLUTION: 6. What is the directrix of the parabola with the equation {{{y+3=(1/10) (x+2)^2}}} ?

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: 6. What is the directrix of the parabola with the equation {{{y+3=(1/10) (x+2)^2}}} ?      Log On


   

Question 33190: 6. What is the directrix of the parabola with the equation y%2B3=%281%2F10%29+%28x%2B2%29%5E2 ?
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
y%2B3=%281%2F10%29+%28x%2B2%29%5E2
STANDARD EQN.OF PARABOLA IS
(X-H)^2=4A(Y-K)...WHERE
FOCUS IS (H,K+A)
DIRECTRIX IS Y-K+A=0
AXIS IS X-H=0
VERTEX IS (H,K)
HERE WE HAVE
(X+2)^2=10(Y+3)...SO COMPARING WE GET
H=-2.....K=-3.....4A=10...OR...A=2.5...HENCE
DIRECTRIX IS GIVEN BY
Y-K+A=0....OR.....
Y-(-3)+2.5=0
Y+3+2.5=0
Y+5.5=0
2Y+11=0