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Math Help - [SOLVED] Straight lines(coordinate geometry)?

  1. #1
    Super Member fardeen_gen's Avatar
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    [SOLVED] Straight lines(coordinate geometry)?

    If p and q are the lengths of the perpendiculars from the origin to the lines x cos θ - y sin θ = k cos 2θ and x sec θ + y cosec θ = k , respectively, prove that p^2 + 4q^2 = k^2.
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    The equation of the line \mathcal{L}: ax+by=c can be written \overrightarrow{OM}\cdot\vec{u}=c, where \vec{u}={a\choose b} (and M is the point we are describing). Then \vec{u} is a vector orthogonal to the line (if M,N\in\mathcal{L}, then \overrightarrow{MN}\cdot\vec{u}=\overrightarrow{ON  }\cdot\vec{u}-\overrightarrow{OM}\cdot\vec{u}=c-c=0), hence the projection of O on \mathcal{L} is the point P such that \overrightarrow{OP}=\lambda\vec{u} and P\in\mathcal{L}, i.e. \lambda\vec{u}\cdot\vec{u}=c, so \lambda=\frac{c}{|\vec{u}|^2} and \overrightarrow{OP}=\frac{c}{|\vec{u}|^2}\vec{u}, and the distance between the origin and the line \mathcal{L} is d=OP=|\overrightarrow{OP}|=\frac{|c|}{|\vec{u}|}. This is a useful result to remember.

    Using this fact, the exercise is just the verification of a simple trigonometric formula. It is even easier if you rewrite the second equation as x\cos\theta+y\sin\theta=k\sin\theta\cos\theta (multiplying both sides by \sin\theta\cos\theta), so that |\vec{u}|=1 for both lines, so you have to check that (k\cos2\theta)^2+4(k\sin\theta\cos\theta)^2=k^2. I guess you know how to do this.

    Laurent.
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    Here is a second way.
    Using the standard distance formula we get:
    p = \frac{{\left| { - k\cos \left( {2\phi } \right)} \right|}}{{\sqrt {\sin ^2 \left( \phi  \right) + \cos ^2 \left( \phi  \right)} }} = \left| k \right|\left| {\cos \left( {2\phi } \right)} \right| \Rightarrow \quad p^2  = k^2 \cos ^2 \left( {2\phi } \right)
    and q = \frac{{\left| { - k} \right|}}{{\sqrt {\sec ^2 \left( \phi  \right) + cse^2 \left( \phi  \right)} }} = \left| k \right|\sqrt {\cos ^2 \left( \phi  \right)\sin ^2 \left( \phi  \right)}.
    From that it follows q^2  = k^2 \cos ^2 \left( {\phi } \right)\sin ^2 \left( {\phi } \right) \Rightarrow \quad 4q^2  = k^2 \sin ^2 \left( {2\phi } \right).
    Now you finish.
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  4. #4
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    Quote Originally Posted by Plato View Post
    Here is a second way.
    (actually our ways are pretty close, because my first paragraph only proves the usual distance formula, which I presumed fardeen_gen did not know)
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  5. #5
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    Quote Originally Posted by Laurent View Post
    (actually our ways are pretty close, because my first paragraph only proves the usual distance formula, which I presumed fardeen_gen did not know)
    I think that it is more likely that fardeen_gen does not know vectors to that extent.
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