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    Stuck on this question, seems simple but not sure what to do:

    x= 4cost, y= 3sint

    The curve crosses the x and y axis at A and B respectively. Find the co-ordinates of A and B.

    I know that when it crosses the y axis, x=0 and when it crosses the x axis y=0 but still not sure how to solve.

    Thanks
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    Crosses x at A, hence y=0 for A. So, for A, 0 = 3sint. Workout t from that and substitute it into the x equation.

    For B, x=0, so 0 = 4cost. Solve for t and sub that value into the y equation.
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    (Original post by MJTravers)
    Crosses x at A, hence y=0 for A. So, for A, 0 = 3sint. Workout t from that and substitute it into the x equation.

    For B, x=0, so 0 = 4cost. Solve for t and sub that value into the y equation.
    I thought so but I don't know how to solve 3sint = 0.
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    (Original post by Endless Blue)
    I thought so but I don't know how to solve 3sint = 0.
    if 3sint = 0, then sint = 0. Do arcsin(0) and that gives you 0

    so t=0, then x = 4cos(0) which is 4.
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    (Original post by MJTravers)
    if 3sint = 0, then sint = 0. Do arcsin(0) and that gives you 0

    so t=0, then x = 4cos(0) which is 4.
    Ah, of course. I think I'd forgotten that I'd worked out t, so thought x=0 which didn't fit with the diagram.

    thank you
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    (Original post by MJTravers)
    x
    A quick binomial expansion q:

    how do I go about the equation (1+x/1-x)^2 ?

    Sorry if that's not clear, I cannot use latex whatsoever; it's a fraction which is all squared.

    Thanks, just not sure how to begin it.
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    (Original post by Endless Blue)
    A quick binomial expansion q:

    how do I go about the equation (1+x/1-x)^2 ?

    Sorry if that's not clear, I cannot use latex whatsoever; it's a fraction which is all squared.

    Thanks, just not sure how to begin it.
    So it's ((1+x)/(1-x))^2?

    Bring the (1-x) to the top, so it becomes ((1+x)(1-x)^-1)^2

    Then, because the ^2 power is on the outside of the brackets, you multiply the powers inside the brackets by 2. So it becomes (1+x)^2(1-x)^-2

    Then I'd expand them seperately using binomial theorem and multiply the products. I'm not 100% sure if that's how you do it, but that's how I'd go about it. Seems to make sense. Let me know if it gives you the right answer!
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    This method works.

    I think that an alternative is to say that (1+x)/(1-x) = -(1+x)/(x-1) = -1 + 2/ (1-x)

    So you have (2/(1-x) - 1)^2, which is easy to expand.

    Either method works fine.

    Posted from TSR Mobile
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    (Original post by tysonmaniac)
    This method works.

    I think that an alternative is to say that (1+x)/(1-x) = -(1+x)/(x-1) = -1 + 2/ (1-x)

    So you have (2/(1-x) - 1)^2, which is easy to expand.

    Either method works fine.

    Posted from TSR Mobile
    That looks way easier than the way I went about it. Could you explain how you go from -(1+x)/(x-1) to -1 + 2/(1-x), please? It's probably really simple but I can't get it
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    (Original post by MJTravers)
    That looks way easier than the way I went about it. Could you explain how you go from -(1+x)/(x-1) to -1 + 2/(1-x), please? It's probably really simple but I can't get it
    Sure:

    Start with -(1+x)/(x-1)

    We know that 1+x = x - 1 + 2

    So we write

    -(1+x)/(x-1) = -(x-1+2)/(x-1)

    -(x-1+2)/(x-1) = [-(x-1) - 2] / (x-1)

    [-(X-1) - 2] / (X-1) = -1 - 2/(x-1) = -1 + 2/(1-x)

    Hope that helps
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    Nobody here knows what they're talking about.

    x=4cost, y=3sint

    Express in cartesian form:

    x/4=cost, y/3=sint

    (x/4)^2 + (y/3)^2 =1

    Crosses x-axis: y=0

    Therefore, (x/4)^2 = 1 ==> x^2/16 = 1 ===> A=+/- 4

    Crosses y-axis: x=0

    Therefore, (y/3)^2 = 1 ==> y^2/9 = 1 ===> B=+/- 3
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    (Original post by qwertyabcdef123)
    Nobody here knows what they're talking about.

    x=4cost, y=3sint

    Express in cartesian form:

    x/4=cost, y/3=sint

    (x/4)^2 + (y/3)^2 =1

    Crosses x-axis: y=0

    Therefore, (x/4)^2 = 1 ==> x^2/16 = 1 ===> A=+/- 4

    Crosses y-axis: x=0

    Therefore, (y/3)^2 = 1 ==> y^2/9 = 1 ===> B=+/- 3
    This method is slightly inefficient, but if you find it easier then it is fine. The point however, is that on should be comfortable working with parametric equations without needing to always convert to Cartesian form.
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    (Original post by qwertyabcdef123)
    Nobody here knows what they're talking about.

    x=4cost, y=3sint

    Express in cartesian form:

    x/4=cost, y/3=sint

    (x/4)^2 + (y/3)^2 =1

    Crosses x-axis: y=0

    Therefore, (x/4)^2 = 1 ==> x^2/16 = 1 ===> A=+/- 4

    Crosses y-axis: x=0

    Therefore, (y/3)^2 = 1 ==> y^2/9 = 1 ===> B=+/- 3
    lol'd, there's no need to switch to cartesian form to solve this and anyone that's done more than ~5 parametric equation questions can see that.
    fail.
 
 
 
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