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Question 32122: find the vertex and intercepts for each parabola. sketch the graph
g(x)=x^2+x-6
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
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.
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
Equations/30056: I need to determine the following for these two problems :HOW MANY X-INTERCEPTS THE PARABOLA HAS, and WHETER ITS VERTEX LIES ABOVE OR BELOW OR ON THE X-AXIS.
1.problem
y=x^2-5x+6
2.problem
y=-x^2+2x-1
1 solutions
Answer 16805 by venugopalramana(1167) About Me on 2006-03-13 10:41:52 (Show Source):
I need to determine the following for these two problems :HOW MANY X-INTERCEPTS THE PARABOLA HAS, and WHETER ITS VERTEX LIES ABOVE OR BELOW OR ON THE X-AXIS.
1.problem
y=x^2-5x+6
PUT Y=0 AND SOLVE FOR X TO GET X INTERCEPTS.
X^2-5X+6=0=X^2-2X-3X+6=0=X(X-2)-3(X-2)=0=(X-2)(X-3)=0....X=2 AND 3...
SO THE X INTERCEPTS ARE AT X = 2 AND X = 3
Y=X^2-5X+6={X^2-2*X*5/2+(5/2)^2}-(5/2)^2+6 =(X-2.5)^2 - 0.25
SO THE VERTEX IS AT X=2.5 AND Y=-0.25...THAT IS BELOW X AXIS
2.problem
y=-x^2+2x-1
DOING THE SAME WAY WE GET
Y=-(X-1)^2=0 AND HENCE
X INTERCEPTS ARE X=1
AND VERTEX IS AT X=1 AND Y=0 SO THE VERTEX IS ON THE X AXIS.
Coordinate-system/29860: I am working with parabolas. For this problem I need to
Complete the square
Give the Vertex
Give the Axis
Give the x-intercepts
Give the y-intercepts
Give a point symmetric to the y-intercept
Draw the graph
The problem is y=x^2-2x-3.
Can anyone help me solve this. I am working on it and would like to have something to check my answer with.
1 solutions
Answer 16624 by venugopalramana(1167) About Me on 2006-03-11 07:58:25 (Show Source):
SEE THE FOLLOWING EXAMPLE WHICH IS ALMOST SAME AS YOUR PROBLEM AND TRY.IF STILL IN DIFFICULTY PLEASE COME BACK.
Y=X^2-2X-3=X^2-2*X*1+1^2-1-3=(X-1)^2-1-3=(X-1)^2-4=0
X INTERCEPT IS OBTAINED BY PUTTING Y=0....WE GET
X^2-2X-3=0=X^2-3X+X-3=X(X-3)+1(X-3)=(X-3)(X+1)=0...SO..X=3 OR -1...
Y INTERCEPT IS GOT BY PUTTING X =0...WE GET
Y=0-0-3=-3
SEE THE GRAPH BELOW ......
graph( 500, 500, -10, 20, -20, 20,x^2-2x-3 )
THE AXIS IS X=1 AS YOU CAN SEE FROM THE GRAPH
ONE POINT SYMMETRIC TO Y INTERCEPT??NO...SYMMETRIC TO AXIS IT IS......(-1,3)
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I have to put this equation y=x^2-2x-15 into this form y=a(x-h)^2+k.
MAKE A PERFECT SQUARE USING X^2 AND X TERMS.ADD AND SUBTRACT THE REQUIRED CONSTANT FOR THE PURPOSE.
Y=(X-1)^2-1-15=(X-1)^2-16...COMPARING WITH THE ABOVE
y=a(x-h)^2+k.
WE GET A=1 AND K=-16
I have to find the line of symmetry.
X-1=0 IS THE LINE OF SYMMETRY SINCE ON EITHER SIDE OF X=1,WE GET SYMMETRIC/SAME VALUES FOR Y..AT X=1+2=3..Y IS -12 AND AT X=1-2=-1 ALSO WE GET Y=-12
(h,k)=vertex I think you use complete the square technique.
YA ..THE VERTEX AS YOU SHOULD KNOW NOW IS AT X=1 AND AT X=1 ,Y=-16 .SO (1,-16) IS THE VERTEX
THE GRAPH WILL LOOK LIKE THIS
graph( 500, 500, -10, 20, -20, 20, x^2-2x-15 )
Please help. thanks
Quadratic-relations-and-conic-sections/28184: What is the vertices, foci, and slope of the asymptotes for the hyperbola whose equation is, y^2/16 - x^2/25 =25?
1 solutions
Answer 15970 by venugopalramana(1088) About Me on 2006-03-04 03:12:07 (Show Source):
SEE THE FOLLOWING AND YOU SHOULD BE ABLE TO SOLVE YOUR PROBLEM BY YOUR SELF.....
THE ANSWERS FOR YOUR CASE...H=0...K=0..A=4...B=5....EQN IS OF THE TYPE
(Y-K)^2/B^2-(X-H)^2/A^2=1....
SO VERTICES ARE...(H,(K-B)) AND (H,(K+B)) ...(0,-5) AND (0,5)
FOCI ARE {H,(K-BE)} AND {H,(K+BE)}...WHERE E IS
ECCENTRICITY =SQRT{(A^2+B^2)/B^2}=SQRT((16+25)/25)=SQRT(41/25)
SO FOCI ARE =(0,-5SQRT(41/25) AND (0,5SQRT(41/25)...
OR....(0,-SQRT(41)) AND (0,SQRT(41)
ASYMPTOTES ARE GIVEN BY
Y^2/16-X^2/25=K
25Y^2-16X^2-400K=0
(5Y+4X+A)(5Y-4X+B)=0
SLOPES OF ASYMPTOTES ARE
-4/5 AND 4/5
THE GRAPHS LOOK LIKE THIS
graph( 600, 600, -10, 10, -10, 10, 4*(1+x^2/25)^0.5,-4*(1+x^2/25)^0.5,4*x/5,-4*x/5 )
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What is the vertices, foci, and slope of the asymptote for the hyperbola whose equation is, y^2 - 4x^2 - 2y - 16x + 1 = 0?
(Y^2-2*Y*1+1^2)-{(2X)^2+2*(2X)*4+4^2}-1^2+4^2+1=0
(Y-1)^2-(2X+4)^2=-16
4(X+2)^2-(Y-1)^2=16
{4(X+2)^2}/16-{(Y-1)^2}/16=1
(X+2)^2/2^2-(Y-1)^2/4^2=1...
COMPARING WITH STANDARD EQN.
(X-H)^2/A^2-(Y-K)^2/B^2=1....WE HAVE
VERTICES ARE {(H-A),K} AND {(H+A),K}=(-2-2,1) AND (-2+2,1)=(-4,1) AND (0,1)
FOCI ARE {(H-AE),K} AND {(H+AE),K}...WHERE E IS
ECCENTRICITY =SQRT{(A^2+B^2)/A^2}=SQRT((4+16)/4)=SQRT(5)
SO FOCI ARE =(-2-2SQRT(5),1) AND (-2+2SQRT(5),1)
SLOPE OF ASYMPTOTE IS GIVEN BY DIFFERENTIATION.HAVE YOU BEEN TAUGHT?PLEASE INFORM.I SHALL COME BACK ON HEARING FROM YOU.
or you can take this proposition as proved formula
the pair of asymptotes for a conic is given by the same equation as the conic except for the constant term which has to be found using the condition for the equation to represent a pair of straight lines.
HENCE EQN OF ASYMPTOTES IS GIVEN BY
y^2 - 4x^2 - 2y - 16x + K=0 , WHERE K IS DETERMINED using the condition for the equation to represent a pair of straight lines.
SINCE WE ARE TO FIND ONLY SLOPES ,WE NEED NOT DETERMINE THE CONSTANT BUT ASSUME THAT THIS EQN REPRESENTS A PAIR OF STRAIGHT LINES.SO
y^2 - 4x^2 - 2y - 16x + K=0 = (Y+2X+A)(Y-2X+B)
HENCE SLOPES ARE +2 AND -2
graph( 600, 600, -10, 10, -10, 10, 1-2*((x+2)^2-4)^0.5,1+2*((x+2)^2-4)^0.5,2x+5,-2x-3 )
Quadratic-relations-and-conic-sections/28185: What is the vertices, foci, and slope of the asymptotes for the hyperbola whose equation is, x^2/81 - y^2/36 = 1?
1 solutions
Answer 15969 by venugopalramana(1088) About Me on 2006-03-04 03:09:19 (Show Source):
SEE THE FOLLOWING AND YOU SHOULD BE ABLE TO SOLVE YOUR PROBLEM BY YOUR SELF.....
THE ANSWERS FOR YOUR CASE...H=0...K=0..A=4...B=5....EQN IS OF THE TYPE
(Y-K)^2/B^2-(X-H)^2/A^2=1....
SO VERTICES ARE...(H,(K-B)) AND (H,(K+B)) ...(0,-5) AND (0,5)
FOCI ARE {H,(K-BE)} AND {H,(K+BE)}...WHERE E IS
ECCENTRICITY =SQRT{(A^2+B^2)/B^2}=SQRT((16+25)/25)=SQRT(41/25)
SO FOCI ARE =(0,-5SQRT(41/25) AND (0,5SQRT(41/25)...
OR....(0,-SQRT(41)) AND (0,SQRT(41)
ASYMPTOTES ARE GIVEN BY
Y^2/16-X^2/25=K
25Y^2-16X^2-400K=0
(5Y+4X+A)(5Y-4X+B)=0
SLOPES OF ASYMPTOTES ARE
-4/5 AND 4/5
THE GRAPHS LOOK LIKE THIS
graph( 600, 600, -10, 10, -10, 10, 4*(1+x^2/25)^0.5,-4*(1+x^2/25)^0.5,4*x/5,-4*x/5 )
------------------------------------------
What is the vertices, foci, and slope of the asymptote for the hyperbola whose equation is, y^2 - 4x^2 - 2y - 16x + 1 = 0?
(Y^2-2*Y*1+1^2)-{(2X)^2+2*(2X)*4+4^2}-1^2+4^2+1=0
(Y-1)^2-(2X+4)^2=-16
4(X+2)^2-(Y-1)^2=16
{4(X+2)^2}/16-{(Y-1)^2}/16=1
(X+2)^2/2^2-(Y-1)^2/4^2=1...
COMPARING WITH STANDARD EQN.
(X-H)^2/A^2-(Y-K)^2/B^2=1....WE HAVE
VERTICES ARE {(H-A),K} AND {(H+A),K}=(-2-2,1) AND (-2+2,1)=(-4,1) AND (0,1)
FOCI ARE {(H-AE),K} AND {(H+AE),K}...WHERE E IS
ECCENTRICITY =SQRT{(A^2+B^2)/A^2}=SQRT((4+16)/4)=SQRT(5)
SO FOCI ARE =(-2-2SQRT(5),1) AND (-2+2SQRT(5),1)
SLOPE OF ASYMPTOTE IS GIVEN BY DIFFERENTIATION.HAVE YOU BEEN TAUGHT?PLEASE INFORM.I SHALL COME BACK ON HEARING FROM YOU.
or you can take this proposition as proved formula
the pair of asymptotes for a conic is given by the same equation as the conic except for the constant term which has to be found using the condition for the equation to represent a pair of straight lines.
HENCE EQN OF ASYMPTOTES IS GIVEN BY
y^2 - 4x^2 - 2y - 16x + K=0 , WHERE K IS DETERMINED using the condition for the equation to represent a pair of straight lines.
SINCE WE ARE TO FIND ONLY SLOPES ,WE NEED NOT DETERMINE THE CONSTANT BUT ASSUME THAT THIS EQN REPRESENTS A PAIR OF STRAIGHT LINES.SO
y^2 - 4x^2 - 2y - 16x + K=0 = (Y+2X+A)(Y-2X+B)
HENCE SLOPES ARE +2 AND -2
graph( 600, 600, -10, 10, -10, 10, 1-2*((x+2)^2-4)^0.5,1+2*((x+2)^2-4)^0.5,2x+5,-2x-3 )
Quadratic-relations-and-conic-sections/28183: What is the center, foci, and the lengths of the major and minor axes for the
ellipse, 16x^2 + 25y^2 + 32x - 150y = 159?
1 solutions
Answer 15968 by venugopalramana(1088) About Me on 2006-03-04 03:04:16 (Show Source):
SEE THE FOLLOWING EXAMPLE
What is the center, foci, and the length of the major and minor axes for the ellipse whose equation is, 16x^2 + 25y^2 + 32x - 150y = 159?
STANDARD EQN.OF ELLIPSE IS
(X-H)^2/A^2 + (Y-K)^2/B^2 = 1
WHERE
CENTRE IS (H,K)
ECCENTRICITY = E = {(A^2-B^2)/A^2}^0.5
FOCI ARE (H+AE,K)AND (H-AE,K)
MAJOR AXIS LENGTH = 2A
MINOR AXIS LENGTH = 2B
WE HAVE
16x^2 + 25y^2 + 32x - 150y - 159 =0
{(4X)^2+2*(4X)*4+4^2}-4^2+{(5Y)^2-2*(5Y)*15+15^2}-15^2-159=0
(4X+4)^2 + (5Y-15)^2 = 400
16(X+1)^2 + 25(Y-3)^2 =400...DIVIDING BY 400 THROUGHOUT..
(X+1)^2/25 + (Y-3)^2/16 =1
(X+1)^2/5^2 + (Y-3)^2/4^2 =1
COMPARING WITH ABOVE STANDARD EQN.
CENTRE IS (-1,3)
ECCENTRICITY E IS {(25-16)/25}^0.5=3/5
FOCI ARE (-1+5*3/5 ,3) AND (-1-5*3/5,3)=(2,3) AND (-4,-3)
MAJOR AXIS LENGTH = 2*5=10
MINOR AXIS LENGTH = 2*4=8
Quadratic-relations-and-conic-sections/28182: What is the center, foci, and the lengths of the major and minor axes for the
ellipse, (x-4)^2/16 + (y+1)^2/9 = 1?
1 solutions
Answer 15966 by venugopalramana(1088) About Me on 2006-03-04 03:01:34 (Show Source):
SEE THE FOLLOWING EXAMPLE
What is the center, foci, and the length of the major and minor axes for the ellipse whose equation is, 16x^2 + 25y^2 + 32x - 150y = 159?
STANDARD EQN.OF ELLIPSE IS
(X-H)^2/A^2 + (Y-K)^2/B^2 = 1
WHERE
CENTRE IS (H,K)
ECCENTRICITY = E = {(A^2-B^2)/A^2}^0.5
FOCI ARE (H+AE,K)AND (H-AE,K)
MAJOR AXIS LENGTH = 2A
MINOR AXIS LENGTH = 2B
WE HAVE
16x^2 + 25y^2 + 32x - 150y - 159 =0
{(4X)^2+2*(4X)*4+4^2}-4^2+{(5Y)^2-2*(5Y)*15+15^2}-15^2-159=0
(4X+4)^2 + (5Y-15)^2 = 400
16(X+1)^2 + 25(Y-3)^2 =400...DIVIDING BY 400 THROUGHOUT..
(X+1)^2/25 + (Y-3)^2/16 =1
(X+1)^2/5^2 + (Y-3)^2/4^2 =1
COMPARING WITH ABOVE STANDARD EQN.
CENTRE IS (-1,3)
ECCENTRICITY E IS {(25-16)/25}^0.5=3/5
FOCI ARE (-1+5*3/5 ,3) AND (-1-5*3/5,3)=(2,3) AND (-4,-3)
MAJOR AXIS LENGTH = 2*5=10
MINOR AXIS LENGTH = 2*4=8
Quadratic-relations-and-conic-sections/28745: Hi, In my Algebra II class we are working on Solving Quadratic Equations by graphing. I relize how to work the problems with a plain vertex like in this problem x2+2x-8=0 the vertex is 1 and so the root/answer would be (-4 and 2) but i am unsure of solving the ones with a fraction for the vertex like in this problem x2-5x+4=0 and the vertex is 5/2 could you please help me?
thanks so much,
Chelsea
1 solutions
Answer 15712 by venugopalramana(1088) About Me on 2006-02-28 01:31:49 (Show Source):
Hi, In my Algebra II class we are working on Solving Quadratic Equations by graphing. I relize how to work the problems with a plain vertex like in this problem x2+2x-8=0
(X+1)^2-9=0 = (X+1)^2- (3)^2
the vertex is 1 ....NO....PLEASE CORRECT
THE VERTEX IS AT X=-1 AND Y = -9....
and so the root/answer would be (-4 and 2)...THAT IS -1+3=2 AND -1-3=-4
OK...GOOD..SEE THE GRAPH BELOW...
but i am unsure of solving the ones with a fraction for the vertex like in this problem x2-5x+4=0
(X-5/2)^2 - 9/4 = 0 = (X-5/2)^2 - (3/2)^2
and the vertex is 5/2
CORRECT ..YOU ARE CORRECT HERE..HOW DID YOU MISTAKE IN THE EARLIER CASE?
could you please help me?
SO THE ROOTS ARE 5/2 + 3/2 =8/2 = 4.......OR.....5/2 - 3/2 = 2/2 =1...THAT IS 4 OR 1.
I HOPE YOU GOT THE METHOD..THE ROOTS ARE OBTAINED BY ADDING /SUBTRACTING THE SQUARE ROOT OF CONSTANT TERM FROM THE VERTEX (OR PRECISELY X COORDINATE OF VERTEX)
SEE THE GRAPH BELOW
graph( 600, 600, -10, 10, -10, 10,-1+(x+9)^0.5,-1-(x+9)^0.5,2.5+(x+2.25)^0.5,2.5-(x+2.25)^0.5,-1,2.5 )
Quadratic-relations-and-conic-sections/28744: I have a parabola that I need to find an equation for.
so far all i have is y=negative___x squared+6. I need to fill in the blank.
I know it passes through (neg.4,5) and (7,2)
help me figure out what goes in the blank please
I just cannot remember how to do it out
1 solutions
Answer 15711 by venugopalramana(1088) About Me on 2006-02-28 01:18:17 (Show Source):
I have a parabola that I need to find an equation for.
so far all i have is y=negative___x squared+6. I need to fill in the blank.
I know it passes through (neg.4,5) and (7,2)
help me figure out what goes in the blank please
I just cannot remember how to do it out
Y=-A*X^2+6...IF IT PASSES THROUGH (-4,5)
5= -A *(-4)^2+6 = -16A+6
16A=6-5=1
A=1/16.....
LET US CHECK WHETHER (7,2) LIES ON THAT
2=-(1/16)*7^2+6 IS NOT CORRECT ..SO PLEASE CHECK YOUR DATA
BUT THIS IS THE METHOD OF SOLUTION.SINCE THERE IS ONLY 1 UNKNOWN WE NEED ONLY ONE POINT..THE SECOND POINT IS REDUNDANT..
Quadratic-relations-and-conic-sections/28461: What is the center, foci, and the length of the major and minor axes for the ellipse whose equation is, (x-4)^2/16 + (y+1)^2/9 = 1?
1 solutions
Answer 15710 by venugopalramana(1088) About Me on 2006-02-28 01:11:31 (Show Source):
SEE THE FOLLOWING EXAMPLE AND TRY. IF STILL IN DIFFICULTY PLEASE COME BACK.
------------------------------------------------------------------------------
What is the center, foci, and the length of the major and minor axes for the ellipse whose equation is, 16x^2 + 25y^2 + 32x - 150y = 159?
STANDARD EQN.OF ELLIPSE IS
(X-H)^2/A^2 + (Y-K)^2/B^2 = 1
WHERE
CENTRE IS (H,K)
ECCENTRICITY = E = {(A^2-B^2)/A^2}^0.5
FOCI ARE (H+AE,K)AND (H-AE,K)
MAJOR AXIS LENGTH = 2A
MINOR AXIS LENGTH = 2B
WE HAVE
16x^2 + 25y^2 + 32x - 150y - 159 =0
{(4X)^2+2*(4X)*4+4^2}-4^2+{(5Y)^2-2*(5Y)*15+15^2}-15^2-159=0
(4X+4)^2 + (5Y-15)^2 = 400
16(X+1)^2 + 25(Y-3)^2 =400...DIVIDING BY 400 THROUGHOUT..
(X+1)^2/25 + (Y-3)^2/16 =1
(X+1)^2/5^2 + (Y-3)^2/4^2 =1
COMPARING WITH ABOVE STANDARD EQN.
CENTRE IS (-1,3)
ECCENTRICITY E IS {(25-16)/25}^0.5=3/5
FOCI ARE (-1+5*3/5 ,3) AND (-1-5*3/5,3)=(2,3) AND (-4,-3)
MAJOR AXIS LENGTH = 2*5=10
MINOR AXIS LENGTH = 2*4=8
Quadratic-relations-and-conic-sections/28463: What is the center, foci, and the length of the major and minor axes for the ellipse whose equation is, 16x^2 + 25y^2 + 32x - 150y = 159?
1 solutions
Answer 15709 by venugopalramana(1088) About Me on 2006-02-28 01:03:38 (Show Source):
What is the center, foci, and the length of the major and minor axes for the ellipse whose equation is, 16x^2 + 25y^2 + 32x - 150y = 159?
STANDARD EQN.OF ELLIPSE IS
(X-H)^2/A^2 + (Y-K)^2/B^2 = 1
WHERE
CENTRE IS (H,K)
ECCENTRICITY = E = {(A^2-B^2)/A^2}^0.5
FOCI ARE (H+AE,K)AND (H-AE,K)
MAJOR AXIS LENGTH = 2A
MINOR AXIS LENGTH = 2B
WE HAVE
16x^2 + 25y^2 + 32x - 150y - 159 =0
{(4X)^2+2*(4X)*4+4^2}-4^2+{(5Y)^2-2*(5Y)*15+15^2}-15^2-159=0
(4X+4)^2 + (5Y-15)^2 = 400
16(X+1)^2 + 25(Y-3)^2 =400...DIVIDING BY 400 THROUGHOUT..
(X+1)^2/25 + (Y-3)^2/16 =1
(X+1)^2/5^2 + (Y-3)^2/4^2 =1
COMPARING WITH ABOVE STANDARD EQN.
CENTRE IS (-1,3)
ECCENTRICITY E IS {(25-16)/25}^0.5=3/5
FOCI ARE (-1+5*3/5 ,3) AND (-1-5*3/5,3)=(2,3) AND (-4,-3)
MAJOR AXIS LENGTH = 2*5=10
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.
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