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    Greetings,

    I'm currently struggling with figuring out part iv) of question 15, shown below. I've tried a few different things, to find the equation or at least the gradient of OT, but I keep coming up with nonsensical equations which don't help me at all.

    Could someone please enlighten me? I'd really appreciate a bit of help.

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    Thank you in advance for your time and assistance.
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    (Original post by DeadManProp)
    ....
    You want to determine the coordinates of R,Q,P then show that the required distances are equal.


    To find Q, you want the line OT. To get the line OT, you want the point T.
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    (Original post by RDKGames)
    You want to determine the coordinates of R,Q,P then show that the required distances are equal.


    To find Q, you want the line OT. To get the line OT, you want the point T.
    I've tried to come up with co-ordinates of T, by equating the expression for the curve with the expression for the tangent, as well as rearranging them to get the expression for x or y and substituting it into the tangent/curve expression. But that gives me equations which seem overly complicated and don't really lead to anywhere. I feel like I'm missing some obvious piece of the puzzle here.
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    (Original post by DeadManProp)
    I've tried to come up with co-ordinates of T, by equating the expression for the curve with the expression for the tangent, as well as rearranging them to get the expression for x or y and substituting it into the tangent/curve expression. But that gives me equations which seem overly complicated and don't really lead to anywhere. I feel like I'm missing some obvious piece of the puzzle here.
    Note that T is not a fixed point. It varies. The general coordinates of it are x=\frac{1}{1+t} and y=\frac{1}{(1+t)(1-t)}

    Thus, the gradient OT is given by m= \dfrac{1}{1-t}, and hence the line is y-\dfrac{1}{(1+t)(1-t)} = \dfrac{1}{1-t}(x - \dfrac{1}{1+t})

    When x=\frac{1}{2} , etc....

    While at it, you should also have found that P varies as well, hence the y-coordinate of it is in terms of t.
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    Ahh I see! I totally forgot that you can use the parametric equations as co-ordinates of T.

    Thank you once again,

    All the best
 
 
 
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