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Math Help - Distance from point to line on hyperbola asymptotes problem

  1. #1
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    Distance from point to line on hyperbola asymptotes problem

    Hi,

    I've been struggling with this for a while now and to no avail. Here's the question.

    A hyperbola of the form

    \frac{x^2}{\alpha^2}-\frac{y^2}{\beta^2}=1

    has asymptotes with equations y^2 = m^2 x^2 and passes through the point (a,0). Find the equation of the hyperbola in terms of x, y, a and m.
    A point P on this hyperbola is equidistant from one of its asymptotes and the x-axis. Prove that, for all values of m, P lies on the curve with equation.

    (x^2-y^2)^2 = 4x^2(x^2-a^2)

    I hate these type of questions as I always end up with a pile of messy algebra and ton of screwed up paper. Here's my attempt this time.

    I got the first part ok, the equation of the hyperbola is

    \frac{x^2}{a^2}-\frac{y^2}{m^2 a^2}=1

    For the second part I decided to pick an arbitrary point on the hyperbola and find its distance from the asymptote and the x-axis and set them equal, then eliminate m. So my general point is P = (a\sec\theta, am\tan\theta). Now the distance of this point to the x-axis is simply am\tan\theta. For the moment I chose the point to be in the first quadrant so the equation of the corresponding asymptote is y=mx. Wolfram says that the distance between the point and the asymptote is

    \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} where the line is Ax + By + C = 0

    So for my point and line this becomes

    \frac{ma\sec\theta + ma\tan\theta}{\sqrt{m^2+1}} which I set equal to the distance from the y-axis to obtain

    \frac{ma \sec \theta + ma\tan\theta}{\sqrt{m^2+1}} = ma\sec\theta

    It is here where I have no idea how to eliminate m to get the required equation or even if I've been doing it wrong the whole time.

    Please can someone help, thanks!!

    Stonehambey
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  2. #2
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    Hi

    Quote Originally Posted by Stonehambey View Post
    Wolfram says that the distance between the point and the asymptote is

    \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} where the line is Ax + By + C = 0

    So for my point and line this becomes

    \frac{ma\sec\theta + ma\tan\theta}{\sqrt{m^2+1}} which I set equal to the distance from the y-axis to obtain

    \frac{ma \sec \theta + ma\tan\theta}{\sqrt{m^2+1}} = ma\sec\theta
    Yes but your equation is y = mx or mx - y = 0 therefore A = m and B = -1 which gives \frac{ma\sec\theta - ma\tan\theta}{\sqrt{m^2+1}}

    Then I do not understand why you set it equal to the distance from the y-axis ? It should be the x-axis right ?

    \frac{ma\sec\theta - ma\tan\theta}{\sqrt{m^2+1}} = ma\tan\theta

    You can then eliminate ma and simplify to get m^2 = \frac{1-2\sin \theta}{\sin^2 \theta}

    Then you can calculate \left(x^2-y^2\right)^2 and 4x^2(x^2-a^2) to show that they are equal
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  3. #3
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    I also had trouble with messy algebra, but I got there in the end, using a more geometric/trigonometric approach.

    The asymptote y=mx makes an angle \theta with the x-axis, where m=\tan\theta. If a point is equidistant from these two lines, then it must lie on their bisector, namely the line y=tx, where t = \tan(\theta/2). The formula for tan of twice an angle then tells you that m = \frac{2t}{1-t^2}.

    Therefore m(1-t^2) = 2t. If the point (x,y) lies on the line y=tx then t = \tfrac yx, and so m\bigl(1-\bigl(\tfrac yx\bigr)^2\bigr) = 2\tfrac yx, from which m = \frac{2xy}{x^2-y^2}.

    If the point (x,y) also lies on the hyperbola then \frac{x^2}{a^2} - \frac{y^2}{m^2a^2} = 1, from which m^2(x^2-a^2) = y^2. Substitute the formula for m from the previous paragraph into that, and you get the result.
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  4. #4
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    That's a pretty smart way!

    I got there in the end by saying that the point be (x,y) and using their relationship only at the end, it went something like this

    y = \frac{mx+y}{\sqrt{m^2+1}}
    y^2 m^2 + y^2 = m^2 x^2 + 2mxy + y^2
    m^2(y^2-x^2) = 2mxy

    Now I square again because this remove any ambiguity as to which asymptote we are referring to, so we now have

    m^4 (y^2 - x^2)^2 = 4m^2 x^2 y^2
    m^2 (y^2 - x^2)^2 = 4 x^2 y^2

    Now, I used the fact that x and y lie on a hyperbola to get

    m^2(x^2 - y^2)^2 = 4x^2 m^2 (x^2-a^2)
    (x^2 - y^2)^2 = 4x^2(x^2 - a^2)

    So I got there in the end, but thank you for all your help I don't know what it is about these type of questions (conic section/locus ones) but they seem to knock me dead with insane algebra every single time. Are they generally quite tricky or am I being really dumb! :P

    Stonehambey
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