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Question 43948: Find an equation of the ellipse having the given points as foci and the given number as sum of focal radii.
(0,-5); (0,5); 20
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
find an
equation of the ellipse having the given points as
foci and the given number as sum of focal radii.
(-9,0);(9,0);30
one of these is it: x^2/15 + y^2/12 =1 or x^2/225 +
y^2/144 =1
or x^2/15 - y^2/12 = 1
1 solutions
Answer 22069 by venugopalramana(1898) About Me on
2006-05-03 09:12:20 (Show Source):
find an equation of the elipse having the given points
as foci and the given number as sum of focal radii.
(-9,0);(9,0);30
PLEASE NOTE THAT SUM OF FOCAL RADII=LENGTH OF MAJOR
AXIS =2A =30....HENCE A=15
SEE THE FOLLOWING AND TRY.IF STILL IN DIFFICULTY
PLEASE COME BACK
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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 =1
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
Quadratic-relations-and-conic-sections/30009: 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
Answer 16810 by venugopalramana(1167) About Me 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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