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    Please could someone tell me how to do this question, I have the answers, just dont know how to do it.


    6. A curve is given parametrically by the equations
    x = 5 cos t, y = -2 + 4 sin t, 0 £ t < 2p.

    (a) Find the coordinates of all the points at which C intersects the coordinate axes, giving your answers in surd form where appropriate.
    (4 marks)
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    x = 5 cos t, y = -2 + 4 sin t, 0 £ t < 2p.


    wen x=0

    5cos t = 0

    cos t = 0

    t = whatever

    wen y=0

    4sin t = 2

    sint =0.5

    t = pi / 6 or whatever it is
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    (Original post by BlueAngel)
    Please could someone tell me how to do this question, I have the answers, just dont know how to do it.


    6. A curve is given parametrically by the equations
    x = 5 cos t, y = -2 + 4 sin t, 0 £ t < 2p.

    (a) Find the coordinates of all the points at which C intersects the coordinate axes, giving your answers in surd form where appropriate.
    (4 marks)
    Set x = 0...find values for t and then corresponding values of y.

    Then set y = 0, find values for t and corresponding values of x

    These coordinates you get are the points wehere the curve cuts the axes.

    G
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    (Original post by BlueAngel)
    Please could someone tell me how to do this question, I have the answers, just dont know how to do it.


    6. A curve is given parametrically by the equations
    x = 5 cos t, y = -2 + 4 sin t, 0 £ t < 2p.

    (a) Find the coordinates of all the points at which C intersects the coordinate axes, giving your answers in surd form where appropriate.
    (4 marks)
    x=0 therefore cost = 0 => t = pi/2 or 3pi/2
    y = 2 or -6 giving (0,2) and (0,-6)

    y=0 => sint = 1/2 => t= pi/6 or 5pi/6
    x = (root3)/2 and -(root3/2)

    giving (root3/2,0) AND (-root3/2,0)

    it is an ellipse

    MB
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    (Original post by musicboy)
    x=0 therefore cost = 0 => t = pi/2 or 3pi/2
    y = 2 or -6 giving (0,2) and (0,-6)

    y=0 => sint = 1/2 => t= pi/6 or 5pi/6
    x = (root3)/2 and -(root3/2)

    giving (root3/2,0) AND (-root3/2,0)

    it is an ellipse

    MB
    Tut tut!! BlueAngel hasn't got that far yet!!!!! You're jumping ahead!! lol

    G
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    (Original post by rockindemon)
    x = 5 cos t, y = -2 + 4 sin t, 0 £ t < 2p.


    wen x=0

    5cos t = 0

    cos t = 0

    t = whatever

    wen y=0

    4sin t = 2

    sint =0.5

    t = pi / 6 or whatever it is


    I understand that - when x = 5 cos t t = pi/2, and 3pi/2
    when y = -2 + 4 sin t t = pi/6 etc...


    But thats not the answer, the answer is (0, 2) , (0, -6) , (5root3/2, 0 ) , and (-5root3/2 , 0). Which I dont understand how you get.
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    (Original post by gzftan)
    Tut tut!! BlueAngel hasn't got that far yet!!!!! You're jumping ahead!! lol

    G
    sorry

    MB
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    (Original post by BlueAngel)
    I understand that - when x = 5 cos t t = pi/2, and 3pi/2
    when y = -2 + 4 sin t t = pi/6 etc...


    But thats not the answer, the answer is (0, 2) , (0, -6) , (5root3/2, 0 ) , and (-5root3/2 , 0). Which I dont understand how you get.
    Look at musicboy's post...he's got it all down!

    G
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    Oh my god, I understand now. Thank you everyone.
    That question now looks so easy, Im so thick, Im gonna fail tomoz. :eek:
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    (Original post by BlueAngel)
    I understand that - when x = 5 cos t t = pi/2, and 3pi/2
    when y = -2 + 4 sin t t = pi/6 etc...


    But thats not the answer, the answer is (0, 2) , (0, -6) , (5root3/2, 0 ) , and (-5root3/2 , 0). Which I dont understand how you get.
    yep, sorry I made a silly error. x = 5cost. cos t = root3/2 and -root3/2 so x = (5root3/2, 0 ) , and (-5root3/2 , 0)

    MB
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    not posted before so i'm not sure if my notations are correct....but....

    x=5cost
    y=-2+4sint

    x/5=cost
    y+2/4=sint

    using sin^2 t+cos^2t=1

    (x/5)^2 + (y+2/4)^2=1

    then substitute x=0 and y=0 into this equation and you get
    y=-6 or y=2
    and then x=+5root3/2 or -5root3/2

    hope this helps

    helen x
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    (Original post by Miss Jones)
    not posted before so i'm not sure if my notations are correct....but....

    x=5cost
    y=-2+4sint

    x/5=cost
    y+2/4=sint

    using sin^2 t+cos^2t=1

    (x/5)^2 + (y+2/4)^2=1

    then substitute x=0 and y=0 into this equation and you get
    y=-6 or y=2
    and then x=+5root3/2 or -5root3/2

    hope this helps

    helen x


    Hey that another method, but its a good one. Thanx.
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    erm sorry if im wrong or something but
    i agree that the parametric gives an elipse shape
    but i dont think its in terms of pi

    x = 5 cos t, y = -2 + 4 sin t

    the furthest distance of x radius is 5
    the furthest distance of y radius is 4

    but the centre is (0,-2)

    then for the point of intersection...
    just use x = 0 and y= 0
    and just find it

    like
    0 = 5cost
    and 1/2 = sin t
    use your calculator to find it
    it isnt something hard.

    thats all i can say.
    i hope i aint wrong..cause ive tried using this way in almost all of the past year questions and got most of it right
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    (Original post by MalaysianDude)
    erm sorry if im wrong or something but
    i agree that the parametric gives an elipse shape
    but i dont think its in terms of pi

    x = 5 cos t, y = -2 + 4 sin t

    the furthest distance of x radius is 5
    the furthest distance of y radius is 4

    but the centre is (0,-2)

    thats all i can say.
    i hope i aint wrong..cause ive tried using this way in almost all of the past year questions and got most of it right

    It is in terms of pi, but you are right the centre is (0,-2). Because the next question asked for a sketch of the ellipse. And the centre is (0, -2) and the radiuses are 5 and 4.
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    then just let x as 0


    find the value of t and put it into the other equation to get the y value

    then repeat it but this time use y as 0
 
 
 
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