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    please help I'm not getting a quadratic...

    The line with equation y = mx - 2 meets the ellips with equation x2 + 4y2 = 16 at the points P and Q.

    Find the coordinates of the mid point M of PQ giving each coordinate in terms of m

    As m varies find, in cartesian form, an equation of the locus of M.

    (Exam style paper in book q6)

    Cheers
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    What I did was to substitute (mx-2) as y into the equation of the ellipse, giving x^2 + 4(mx-2)^2 = 16.

    I then solved this for x, and found y by doing mx-2.

    The second part's simply converting a parametric equation with parameter m to a cartesian equation.
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    yeah i got that far but can't solve for x in the quadratic!
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    OK, here's my working:

    x² + 4(mx - 2)² = 16
    x² + 4(m²x² - 4mx + 4) = 16
    x² + 4m²x² - 16mx = 0
    (4m² + 1)x² = 16mx Divide through by x (noting that x could be 0)
    (4m² + 1)x = 16m
    so x = 16m/(4m² + 1)
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    that's what i got, but shouldn't i have two answers for the points P AND Q??!! ah im getting so frustrated with P5!!!
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    (Original post by Lucky Penny)
    that's what i got, but shouldn't i have two answers for the points P AND Q??!! ah im getting so frustrated with P5!!!
    Yes there are - one is the point (0,-2) and the other has just been calculated above
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    sorry this aint actually p5 , its p3 but didnt wanna start a new thread for it
    can anyone tell me how to sketch parametric curves :confused:
    in june01 (aqa) it says....

    1a) Sketch the curve given by the paramentric equations (1mark :eek: )

    x=4t² , y=8t for t>0 (greater than or equal to)

    thanks for any help
    Sumit
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    (Original post by sumitk87)
    sorry this aint actually p5 , its p3 but didnt wanna start a new thread for it
    can anyone tell me how to sketch parametric curves :confused:
    in june01 (aqa) it says....

    1a) Sketch the curve given by the paramentric equations (1mark :eek: )

    x=4t² , y=8t for t>0 (greater than or equal to)

    thanks for any help
    Sumit
    You could spot the trend that as t increases for 0<t<1, y increases much more than x while as t increases for t>1, x increases much more than y so it levels off.
    You could alternatively find the cartesian equation [t^2=x/4, t^2=y^2/64] x/4=y^2/64 so y^2=16x and use that to spot the relationship. [y^2=16x also shows that the curve is symmetrical about the x axis, ie: if (a,b) lies on the curve so does (a,-b)

    Usually when analysing curves you look at the following key areas:
    When does the curve touch the axes?
    What happens as the variable tends to 0.
    What happens as the variable tends to +- ∞
    Does the curve have any symmetry?
    As one variable increases, does the other variable increase more quickly or more slowly?
    Are there any undefined points where there might be asymptotes etc?
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    (Original post by sumitk87)
    sorry this aint actually p5 , its p3 but didnt wanna start a new thread for it
    can anyone tell me how to sketch parametric curves :confused:
    in june01 (aqa) it says....

    1a) Sketch the curve given by the paramentric equations (1mark :eek: )

    x=4t² , y=8t for t>0 (greater than or equal to)

    thanks for any help
    Sumit
    If you eliminate t you see that

    y = 8 sqrt(x/4) = 4 sqrt(x) [as y is positive]

    It's just half of a parabola (on its side compared with y = x^2 so that it make the top half of a C rather than a U shape)
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    (Original post by Gaz031)
    You could spot the trend that as t increases for 0<t<1, y increases much more than x while as t increases for t>1, x increases much more than y so it levels off.
    You could alternatively find the cartesian equation [t^2=x/4, t^2=y^2/64] x/4=y^2/64 so y^2=16x and use that to spot the relationship.
    thanks for that but i still think for 1 mark its to much
    but thanks anyway
 
 
 
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