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    The rectangular hyperbola H has equation  xy = c^2 , where c is a constant.
    The point  P(ct,\frac {c}{t}) is a general point on H.

    a) Show that the tangent to H at P has equation

     t^2y + x = 2ct

    The tangents to H at the points A and B meet at the point  (15c, -c)

    b) Find, in terms of c, the coordinates of A and B

    I'm only struggling with b)

    The solution subs the x & y coordinate of the point (15c, -c) into  t^2y + x = 2ct and cancels all the c's and has an equation for t and magically finds the points A & B. I would like to know why/how this was possible.

    My thought process is of the following: We found the tangent at the general point therefore there must be some correlation as to how we are able to just sub the coordinates into the tangent.

    Here is the solution: (Question7)
    https://7cba9babeb0db0ff9468853e0b2d...%20Edexcel.pdf
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    (Original post by Psst.)
    The rectangular hyperbola H has equation  xy = c^2 , where c is a constant.
    The point  P(ct,\frac {c}{t} is a general point on H.

    a) Show that the tangent to H at P has equation

     t^2y + x = 2ct

    The tangents to H at the points A and B meet at the point  (15c, -c)

    b) Find, in terms of c, the coordinates of A and B

    I'm only struggling with b)

    The solution subs the x & y coordinate of the point (15c, -c) into  t^2y + x = 2ct and cancels all the c's and has an equation for t and magically finds the points A & B. I would like to know why/how this was possible.

    My thought process is of the following: We found the tangent at the general point therefore there must be some correlation as to how we are able to just sub the coordinates into the tangent.

    Here is the solution: (Question7)
    https://7cba9babeb0db0ff9468853e0b2d...%20Edexcel.pdf
    They are definitely doing correct ...
    Gosh I am too tired to explain ...
    Where is everybody?
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    (Original post by Psst.)
    The rectangular hyperbola H has equation  xy = c^2 , where c is a constant.
    The point  P(ct,\frac {c}{t}) is a general point on H.

    a) Show that the tangent to H at P has equation

     t^2y + x = 2ct

    The tangents to H at the points A and B meet at the point  (15c, -c)

    b) Find, in terms of c, the coordinates of A and B

    I'm only struggling with b)

    The solution subs the x & y coordinate of the point (15c, -c) into  t^2y + x = 2ct and cancels all the c's and has an equation for t and magically finds the points A & B. I would like to know why/how this was possible.

    My thought process is of the following: We found the tangent at the general point therefore there must be some correlation as to how we are able to just sub the coordinates into the tangent.

    Here is the solution: (Question7)
    https://7cba9babeb0db0ff9468853e0b2d...%20Edexcel.pdf
    do you know what t actually does in this question?


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    (Original post by Psst.)
    The rectangular hyperbola H has equation  xy = c^2 , where c is a constant.
    The point  P(ct,\frac {c}{t}) is a general point on H.

    a) Show that the tangent to H at P has equation

     t^2y + x = 2ct

    The tangents to H at the points A and B meet at the point  (15c, -c)

    b) Find, in terms of c, the coordinates of A and B

    I'm only struggling with b)

    The solution subs the x & y coordinate of the point (15c, -c) into  t^2y + x = 2ct and cancels all the c's and has an equation for t and magically finds the points A & B. I would like to know why/how this was possible.

    My thought process is of the following: We found the tangent at the general point therefore there must be some correlation as to how we are able to just sub the coordinates into the tangent.

    Here is the solution: (Question7)
    https://7cba9babeb0db0ff9468853e0b2d...%20Edexcel.pdf
    They meet at (15c, -c) so you know that when x=15c, y=-c. Substituting those values allows you to find t.
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    (Original post by Psst.)
    The rectangular hyperbola H has equation  xy = c^2 , where c is a constant.
    The point  P(ct,\frac {c}{t}) is a general point on H.

    a) Show that the tangent to H at P has equation

     t^2y + x = 2ct

    The tangents to H at the points A and B meet at the point  (15c, -c)

    b) Find, in terms of c, the coordinates of A and B

    I'm only struggling with b)

    The solution subs the x & y coordinate of the point (15c, -c) into  t^2y + x = 2ct and cancels all the c's and has an equation for t and magically finds the points A & B. I would like to know why/how this was possible.

    My thought process is of the following: We found the tangent at the general point therefore there must be some correlation as to how we are able to just sub the coordinates into the tangent.

    Here is the solution: (Question7)
    https://7cba9babeb0db0ff9468853e0b2d...%20Edexcel.pdf
    Since P represents any point on the curve, the tangent at P represents any tangent. If you enter certain values of x and y (into the equation for the "general tangent") you'll be able to find the corresponding values of c and t (which are used to express any point P on the curve). In this case the values of x and y (at the intersection point) correspond to the values of c and t that make up the coordinates of A and B.

    Hope that sheds some light on it!
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    (Original post by physicsmaths)
    do you know what t actually does in this question?


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    Yes, I think. t is a parameter. solving the quadratic equation will find the values of t for A & B. But what I don't understand is why they subbed in the point (15c,-c) in the tangent? How was they able to do this?
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    Understood now thanks all!
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    (Original post by Psst.)
    Yes, I think. t is a parameter. solving the quadratic equation will find the values of t for A & B. But what I don't understand is why they subbed in the point (15c,-c) in the tangent? How was they able to do this?
    The tangent equation is a function which outputs coordinates like (15c,-c). Recall that y=x^2, for example, will have coordinates like (0,0) and (2,4) lying on it. If you were told that y=ax^2 and were given the coordinate (2,8) then you'd be able to sub in and determine that a=2.
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    (Original post by Psst.)
    The rectangular hyperbola H has equation  xy = c^2 , where c is a constant.
    The point  P(ct,\frac {c}{t}) is a general point on H.

    a) Show that the tangent to H at P has equation

     t^2y + x = 2ct

    The tangents to H at the points A and B meet at the point  (15c, -c)

    b) Find, in terms of c, the coordinates of A and B

    I'm only struggling with b)

    The solution subs the x & y coordinate of the point (15c, -c) into  t^2y + x = 2ct and cancels all the c's and has an equation for t and magically finds the points A & B. I would like to know why/how this was possible.

    My thought process is of the following: We found the tangent at the general point therefore there must be some correlation as to how we are able to just sub the coordinates into the tangent.

    Here is the solution: (Question7)
    https://7cba9babeb0db0ff9468853e0b2d...%20Edexcel.pdf
    This might help;

    Edit: too late, damn mobile :P

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    (Original post by DylanJ42)
    This might help;

    Edit: too late, damn mobile :P

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    No worries, thank you. I'll be using this little fact to destroy other questions, so it's all good. It's always good to find out why you're doing something, instead of just applying recipes to questions that look similar to one another
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    (Original post by Psst.)
    No worries, thank you. I'll be using this little fact to destroy other questions, so it's all good. It's always good to find out why you're doing something, instead of just applying recipes to questions that look similar to one another
    Couldn't agree more :closedeyes:, it's great that you want to understand the content rather than just learn it all off
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    (Original post by Psst.)
    Yes, I think. t is a parameter. solving the quadratic equation will find the values of t for A & B. But what I don't understand is why they subbed in the point (15c,-c) in the tangent? How was they able to do this?
    Because it lies on both tangents for two some values of t.


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