SOLUTION: Find the vertex, focus, and directrix of the parabola and graph it. x^2-2x+8y+9=0.

Question 34716This question is from textbook College Algebra
: Find the vertex, focus, and directrix of the parabola and graph it.
x^2-2x+8y+9=0. This question is from textbook College Algebra

Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
SEE THE FOLLOWING AND TRY
Quadratic-relations-and-conic-sections/33190: 6. What is the directrix of the parabola with the equation y+3=(1/10) (x+2)^2 ?
1 solutions
Answer 19633 by venugopalramana(1377) About Me on 2006-04-11 00:48:42 (Show Source):
y+3=(1/10) (x+2)^2
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
Linear_Algebra/30362: Question: Find the equation of the ellipse whose center is (5,-3) that has a vertex at 13,-3) and a minor axis of lenght 10.
POssible Answers:
(A) (x-5)^2/64 + (y+3)^2/25 = 1
(B) (x+5)^2/64 + (y-3)^2/25 = 1
(C) x^2/64 + y^2/25 = 1
(D) none of these
1 solutions
Answer 17014 by venugopalramana(1167) About Me on 2006-03-15 11:21:03 (Show Source):
SEE THE FOLLOWING AND TRY..IF STILL IN DIFFICULTY PLEASE COME BACK...
OK I WORKED IT OUT FOR YOU NOW
I TOLD YOU EQN IS
(X-H)^2/A^2 + (Y-K)^2/B^2....
WHERE H,K IS CENTRE...SO H=5 AND K=-3 AS CENTRE IS GIVEN AS (5,-3)....NOW VERTEX IS (13,-3)...IT LIES ON ELLIPSE..SO IT SATISFIES THE EQN
(13-5)^2/A^2 +(-3+3)^2/B^2 =1
HENCE A^2=64...OR A=8
MINOR AXIS =10=2B...HENCE B=5..SO EQN.S
(X-H)^2/64 + (Y+3)^2/25 =1
THAT IS A IS CORRECT.
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