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Math Help - question of standard form for an ellipse

  1. #1
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    question of standard form for an ellipse

    Hello, so basically I have previously taken applied calculus about 2 years ago. Friend is now taking precalc and asked for help. Needless to say..ive forgotten much of what I learned. So before I go on helping..have some problems I am attempting that I want to check myself on. So I will probably update this with various questions to check myself if that's ok. Thanks

    First up, putting equation into standard form for a conic. (are all these ellipses?)

    3x^2-y^2+6x+6y - 10 = 0

    So complete the square I ended up with

    3(x+1)^2-(y+3)^2 = 20

    putting in standard form by dividing by 20 I came up with

    [3(X+1)^2]/20-[(y+3)^2]/20 = 1


    is this answer correct? Thanks

    and 2nd

    4x^2+9y^2+24x+18y+9 = 0

    4(x+3)^2=9(y+1)^2 = 1******-9 + (1^2)+(3^2) = 1

    since already = 1..is in standard form?

    last question for standard form

    x^2+10x = -2y+1

    x^2+10x+2y = 1
    (x+5)^2+2y = 26

    [((x+5)^2)/26]+(2y/26) = 1
    Last edited by RYdis; October 17th 2011 at 08:36 PM.
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  2. #2
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    Re: question of standard form for an ellipse

    The first one is an ellipse centered at (x,y)=(-1,-3) with semi-major axis of √(20), semi-minor axis of √(20/3) .

    Standard Form: \frac{(x+1)^2}{20/3}-\frac{(y+3)^2}{20} = 1

    The second one is a hyperbola:

    Standard Form: \frac{(x+3)^2}{4}-\frac{(y+1)^2}{9} = 1

    The third is a parabola:

    y =  -(1/2)(x+5)^2 + 13
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  3. #3
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    Re: question of standard form for an ellipse

    1) This is not an ellipse.

    You can tell what a conic will be by looking at A and B in the equation Ax^2 + By^2 +.... = 0
    A = B --> circle
    A or B = 0 --> parabola
    A, B have same sign --> ellipse
    A, B have different sign --> hyperbola.

    We have a HYPERBOLA (or a typo...)

    2) Let's clear up 1) before proceeding
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    Re: question of standard form for an ellipse

    Quote Originally Posted by TheChaz View Post
    1) This is not an ellipse.

    You can tell what a conic will be by looking at A and B in the equation Ax^2 + By^2 +.... = 0
    A = B --> circle
    A or B = 0 --> parabola
    A, B have same sign --> ellipse
    A, B have different sign --> hyperbola.

    We have a HYPERBOLA (or a typo...)

    2) Let's clear up 1) before proceeding
    your right, I did copy the wrong question. Had it all confused as I wrote this on my paper lol. That first #1 was just to put into standard conic form. After reading the more into the book found out how to tell the difference.

    So new #1.

    4(x+2)^2+6(y-3)^2 = 24

    came up with ((x+2)^2/6) + ((y-3)^2)/4 = 1

    Center (-2,3) (being reversed because standard form is (x-h) and (y-k) correct?
    c^2 = 20. (foci distance)
    vertices being (-2, 3+sqrt(6)) and (-2, 3-sqrt(6))

    do I have these correct?

    still looking for foci and minor vertices
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  5. #5
    Super Member TheChaz's Avatar
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    Re: question of standard form for an ellipse

    Quote Originally Posted by RYdis View Post
    ...
    So new #1.

    4(x+2)^2+6(y-3)^2 = 24

    came up with ((x+2)^2/6) + ((y-3)^2)/4 = 1

    Center (-2,3) (being reversed because standard form is (x-h) and (y-k) correct? [COLOR="rgb(0, 0, 0)"]Correct. [/COLOR]
    c^2 = 20. (foci distance) Where did you get this? A^2 = 6 and B^2 = 4, so...
    vertices being (-2, 3+sqrt(6)) and (-2, 3-sqrt(6))

    do I have these correct?

    still looking for foci and minor vertices
    (See above)
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  6. #6
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    Re: question of standard form for an ellipse

    Quote Originally Posted by TheChaz View Post
    (See above)
    divided by 24 on both sides and reduced the whole numbers to get one on the right side. I assumed that since the bottom is always a^2 and b^2 that getting the actual number would be square rooting it (which is why the vertices would be sqrt(6)?)

    4/24 and 6/24
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  7. #7
    Super Member TheChaz's Avatar
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    Re: question of standard form for an ellipse

    I meant, specifically, where did you get c^2 = 20 !

    A^2 = 6
    B^2 = 4
    C^2 = A^2 - B^2 = 2.
    So C = √2.
    Then go left and right √2 units from (-2, 3) to get (-2 - √2, 3) and (-2 + √2, 3) for your foci.
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  8. #8
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    Re: question of standard form for an ellipse

    Quote Originally Posted by TheChaz View Post
    I meant, specifically, where did you get c^2 = 20 !

    A^2 = 6
    B^2 = 4
    C^2 = A^2 - B^2 = 2.
    So C = √2.
    Then go left and right √2 units from (-2, 3) to get (-2 - √2, 3) and (-2 + √2, 3) for your foci.
    ah, I must of squared 6 and 4 36-16.

    so I can do center, major vertices and foci..working on just minor then.
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  9. #9
    Super Member TheChaz's Avatar
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    Re: question of standard form for an ellipse

    The endpoints of the minor axis are "b" units from the center.
    So go up and down √4 = 2 units from the center.
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    Re: question of standard form for an ellipse

    so the semi vertices are b, in this case 2.

    When determining the vertices, how do you determine when when you should move on the x or y axis?

    if going with wider than taller (a^2 going under x and b^2 under y) you just associate it this way? since b is under the y section, you move on the y axis? So if it was taller than wider, and B^2 was under x, then b^2 moves along the x axis?
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  11. #11
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    Re: question of standard form and conics

    alright, lets see if I can do better. Moving on

    16x^2-4y^2 = 64. So Hyperbola

    (x^2/4)-(y^2/16) = 1
    C^2 = 4+16 = 20

    So Center is (0,0)
    since x-y then vertices (2,0)(-2,0)
    Foci (0, sqrt(20)) (0,-sqrt(20))
    asym = y= +/- 2x

    I have this one correct?

    and #3

    (y-2)^2 = 10(x+1) = parabola
    4p = 10 p = 5/2

    Vertex (-1,2)
    Focus (-1+5/2, 2)
    Dirextrix x= -1-5/2
    Axis of Sym y=2

    Right now just want to check my work. I dont have an answer sheet for these problems, so just making sure im doing right
    Last edited by RYdis; October 18th 2011 at 05:03 PM.
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  12. #12
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    Re: question of standard form and conics

    trouble factoring this equation..really stumped..can't get come out even :/

    3x^2-y^2+6x+6y-10 = 0

    3x^2+6x-y^2+6y = 10

    3(x^2+2x)-(y^2+6y) = 10
    3(x^2+2x+1) - (y^2+6y+9) = 10+9
    3(x+1)^2=3(1) - (y+3)^2 = 19+3..but 3 doesn't divide evenly into 22.. like wise..if I just subtract out the 1..doesn't divide evenly into 20 :/
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