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Conics Hyperbola

  1. Aug 22, 2009 #1

    Mentallic

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    1. The problem statement, all variables and given/known data
    http://img11.imageshack.us/img11/6340/conicshyperbola1.jpg [Broken]

    2. Relevant equations
    [tex]d^2=(x_2-x_1)^2+(y_2-y_1)^2[/tex]

    [tex]y-y_1=m(x-x_1)[/tex]

    [tex]m_1m_2=-1[/tex]


    3. The attempt at a solution
    I was able to answer (i) but for (ii) I would go about it like this:

    Find the equation of the line SR by using that it is perpendicular to the line l and passing through the focus S(ae,0). Then solve both equations simultaneously to find the point of intersection at R(x,y). Then find the distance between S and R given that I know both coordinates.
    But looking at the marking criteria, it is only worth 1 mark and thus must have a much simpler way of being solved. Any ideas?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Aug 23, 2009 #2

    Tom Mattson

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    The only other thing I can see is to let point [itex]Q[/itex] be the x-intercept of the line [itex]l[/itex] and let [itex]R=(x,y)[/itex]. Then you can apply the Pythagorean Theorem to triangle [itex]QRS[/itex] and solve for [itex](x,y)[/itex]. The nice thing about this is that both points [itex]Q[/itex] and [itex]S[/itex] have only one nonzero coordinate.
     
  4. Aug 24, 2009 #3

    Mentallic

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    Well then [tex]Q(acos\theta,0)[/tex] but the problem with that technique is that I'll still need to use the distance formula after finding point R, which is basically not any faster than what my original thinking was.

    I asked my teacher on this one, there is a basic formula that I completely forgot existed.

    [tex]d=\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}[/tex]

    gives the perpendicular distance from a point to a line.
     
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