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    Attached. Done part A by differentiating via chain rule, and getting values by substituting t. got 2/3 and -2/3.

    part b used my answer to part a to form equations in form y-y1=m(x-x1)

    Answer isn't the same line though...

    Could someone also walk me through how you would work out part c?

    you know it meets y axis therefore x =0. sub x=0 into equation of tangent. However could I work this out by just subtituting values of t into original equations then I will know the coordinates and can work out the area from there?
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    (Original post by Xphoenix)
    Attached. Done part A by differentiating via chain rule, and getting values by substituting t. got 2/3 and -2/3.

    part b used my answer to part a to form equations in form y-y1=m(x-x1)

    Answer isn't the same line though...

    I get y=-\frac{2}{3}x+1 for both lines.

    Could someone also walk me through how you would work out part c?

    you know it meets y axis therefore x =0. sub x=0 into equation of tangent. However could I work this out by just subtituting values of t into original equations then I will know the coordinates and can work out the area from there?
    Bit puzzled by this. Question says tangent meets the y-axis at Q, but Q is one of our points and it's not on the y-axis. So, it's not clear which triangle they're talking about.
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    (Original post by ghostwalker)
    I get y=-\frac{2}{3}x+1 for both lines.



    Bit puzzled by this. Question says tangent meets the y-axis at Q, but Q is one of our points and it's not on the y-axis. So, it's not clear which triangle they're talking about.
    Can u show it at P for part b? since that's the one that doesn't have the answer. I'm having a problem with the root 2, cause it's the 'wrong sign' however mathematically it isn't in my method.
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    (Original post by Xphoenix)
    Can u show it at P for part b? since that's the one that doesn't have the answer. I'm having a problem with the root 2, cause it's the 'wrong sign' however mathematically it isn't in my method.
    At P, t=pi/4, and eqn of tgt is:

    y-(1-\frac{2}{\sqrt{2}})=-\frac{2}{3}(x-\frac{3}{\sqrt{2}})
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    (Original post by ghostwalker)
    At P, t=pi/4, and eqn of tgt is:

    y-(1-\frac{2}{\sqrt{2}})=-\frac{2}{3}(x-\frac{3}{\sqrt{2}})
    oh sorry, meant Q then, whichever one where t=5pi/4. I got that one and got it to equal y=-2/3 x +1
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    (Original post by Xphoenix)
    oh sorry, meant Q then, whichever one where t=5pi/4. I got that one and got it to equal y=-2/3 x +1

    At Q, t=5pi/4, and eqn of tgt is:

    y-(1-(-\frac{2}{\sqrt{2}}))=-\frac{2}{3}(x-(-\frac{3}{\sqrt{2}}))
 
 
 
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