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Math Help - Locus/Hyperbola question

  1. #1
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    Locus/Hyperbola question

    Hey guys,

    I'm really bad at these types of questions, I don't know what it is about them but they always seem to stump me. Here it is,

    The variable point P(a\sec t, b\tan t) is on the hyperbola with equation \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 and N is the point (3a, 3b). The point Q lies on the line PN so that PQ = 2QN. As t varies find, in cartesian form, the locus of Q.

    I tried to find the line PN by finding its gradient by

    \frac{3b - b\tan t}{3a - a\sec t}

    and then using the usual method for finding the equation of a straight line, that is

    \frac{y-3b}{x-3a} = \frac{3b - b\tan t}{3a - a\sec t}

    but then I get stuck and cannot see what to do next. Do I need to find the point Q? I know I need to eliminate t, but I just can't see how to approach this one.

    Are there any general tips and tricks common to conic section questions like this? Given a formula I can identify which conic it is, and its directrix, foci etc. But when the questions get slightly harder (like this one) I just go to pieces.

    Any help would be much appreciated

    Stonehambey
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  2. #2
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    Hi

    When you say that the point Q lies on the line PN so that PQ = 2QN, do you mean that Q is between P and N ? Otherwise there are 2 such points

    P------------Q1-------N-------------------Q2
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  3. #3
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    Quote Originally Posted by running-gag View Post
    Hi

    When you say that the point Q lies on the line PN so that PQ = 2QN, do you mean that Q is between P and N ? Otherwise there are 2 such points

    P------------Q1-------N-------------------Q2
    "The point Q lies on the line PN so that \vec {PQ} = 2\vec{QN}"

    That is the quoted from the question verbatim
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  4. #4
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    OK

    Actually you don't need to find a Cartesian equation of line NP

    From \overrightarrow{PQ} = 2\:\overrightarrow{QN} you can find

    \overrightarrow{NQ} = \frac{1}{3}\:\overrightarrow{NP}

    from which you can find the coordinates of Q and then eliminate t

    Spoiler:

    \frac{9(x_Q-2a)^2}{a^2} - \frac{9(y_Q-2b)^2}{b^2} = 1

    \frac{(x_Q-2a)^2}{\left(\frac{a}{3}\right)^2} - \frac{(y_Q-2b)^2}{\left(\frac{b}{3}\right)^2} = 1

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  5. #5
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    ok thanks, I will give this question another go when I get back from work this evening
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  6. #6
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    I'm afraid I'm still having trouble with this one

    Do I need to find the lengths of NQ and NP using the pythagorean theorem? Then by setting NQ = (1/3)NP will eliminate t?

    Thanks again
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  7. #7
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    Sorry, could you better explain where you get stuck ?
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  8. #8
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    Quote Originally Posted by running-gag View Post
    Sorry, could you better explain where you get stuck ?
    Sure,

    Well as you say we know that \overrightarrow{NQ} = \frac{1}{3}\:\overrightarrow{NP}<br />

    So does this mean I just have to work out the lengths of \overrightarrow{NP} and \overrightarrow{NQ} and continue the problem that way? I really don't see the way forward and I tend to write out pages of algebra on these types of problems, so I just wanna know I'm going in the right direction

    I'm sorry, I have a whole host of conic section problems I can't do. It seems to be a complete brick wall for me. I just can't seem to answer questions on them no matter how much I practice
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  9. #9
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    Hello Stonehambey
    Quote Originally Posted by Stonehambey View Post
    Sure,

    Well as you say we know that \overrightarrow{NQ} = \frac{1}{3}\:\overrightarrow{NP}<br />

    So does this mean I just have to work out the lengths of \overrightarrow{NP} and \overrightarrow{NQ} and continue the problem that way?
    No. What you need is this:

    If A, B have position vectors \vec{a}, \vec{b} then the position vector \vec{r} of the point R that divides AB internally in the ratio \lambda:\mu is given by:

    \vec{r} = \frac{\mu\vec{a}+\lambda\vec{b}}{\lambda+\mu}

    This is a very useful formula. Learn it!


    The point P has coordinates (a\sec t, b\tan t); the point N has coordinates (3a, 3b), and Q divides PN internally in the ratio 2:1. (That's what \vec{PQ} = 2\vec{QN} means.) So, using the formula above, the coordinates of Q are:

     \Big(\tfrac13(a\sec t + 6a), \tfrac13(b\tan t + 6b)\Big)

    So the locus of Q is given by

    x =\tfrac13(a\sec t + 6a), y=\tfrac13(b\tan t + 6b)

    i.e. \sec t = \frac{3(x-2a)}{a}, \tan t =\frac{3(y-2b)}{b}

    Eliminating t:

    \frac{9(x-2a)^2}{a^2}-\frac{9(y-2b)^2}{b^2}=1

    Grandad
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