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Question 36551: I was wondering if anyone could help me determine what type of figure is given, then graph. Show all work and label all key points (asymptotes, foci, vertices, the directrix, the center) where applicable.
4(x-1)2 = 4-y^2
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! TIP:
IF X^2 AND Y^2 HAVE DIFFERENT COEFFICIENTS OF SAME SIGN ,THEN IT COULD
BE ELLIPSE.
4.) 4(x-1)2 = 4-y2
I HOPE IT IS
4(X-1)^2+Y^2=4...DIVIDE WITH 4
(X-1)^2/1^2+Y^2/2^2=1
THIS THE EQN.OF AN ELLIPSE.STD.EQN. IS
(X-H)^2/A^2 +(Y-K)^2/B^2=1
WHERE
(H,K) IS CENTRTE....(1,0).HERE
LENGTH OF MAJOR AXIS IS 2B...2*2=4
MINOR AXIS IS ALONG Y=K....Y=0
LENGTH OF MINOR AXIS IS 2A...2*1=2
ECCENTRICITY=E=SQRT{(B^2-A^2)/A^2}
=SQRT(4-1)/1=SQRT(3)
B*E=2SQRT(3)
B/E=2/SQRT(3)
FOCI ARE (H,K+BE) AND (H,K-BE)
(1,2SQRT(3)) AND (1,-2SQRT(3))
DIRECTRIX ARE.Y=K+B/E AND K-B/E
Y=2/SQRT(3)..AND...Y=-2/SQRT(3)
GRAPH IS GIVEN BELOW...
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