Question 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
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! 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.
Can you help me write an equation for an ellipse with a major axis with endpoints of (0,8), and (0,-8) with foci of (0,5) and (0,-5)?
1 solutions
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Answer 16810 by venugopalramana(1120) on 2006-03-13 11:19:12 (Show Source):
Can you help me write an equation for an ellipse with a major axis with endpoints of (0,8), and (0,-8) with foci of (0,5) and (0,-5)?
THIS SHOWS THAT X AXIS IS THE MAJOR AXIS
STANDARD EQN.OF ELLIPSE IS
(X-H)^2/A^2 +(Y-K)^2/B^2=1
CENTRE IS (H,K)..AS PER THE PROBLEM H=K=0 AS CENTRE OF ELLIPSE IS AT (0,0)..SINCE major axis with endpoints ARE (0,8), and (0,-8)
WHERE MAJOR AXIS =2A=8+8=16...SO A=8..SINCE major axis with endpoints ARE (0,8), and (0,-8)
FOCI ARE GIVEN BY
AE,0 AND -AE,0...SO AE =5...SO E=5/A=5/8
BUT E=SQRT{(A^2-B^2)/A^2}=5/8...SQUARING
25/64=(A^2-B^2)/A^2=1-B^2/A^2
B^2/64=1-25/64=49/64
B^2=49
B=7
HENCE EQN. OF ELLIPSE IS
X^2/64 + Y^2/49 = 1
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