Results 1 to 4 of 4

Thread: Distance from point to line on hyperbola asymptotes problem

  1. #1
    Member
    Joined
    Nov 2008
    Posts
    114

    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

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

    has asymptotes with equations $\displaystyle y^2 = m^2 x^2$ 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.

    $\displaystyle (x^2-y^2)^2 = 4x^2(x^2-a^2)$

    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

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

    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 $\displaystyle P = (a\sec\theta, am\tan\theta)$. Now the distance of this point to the x-axis is simply $\displaystyle am\tan\theta$. For the moment I chose the point to be in the first quadrant so the equation of the corresponding asymptote is $\displaystyle y=mx$. Wolfram says that the distance between the point and the asymptote is

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

    So for my point and line this becomes

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

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

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2008
    From
    France
    Posts
    1,458
    Hi

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

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

    So for my point and line this becomes

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

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

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

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

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

    Then you can calculate $\displaystyle \left(x^2-y^2\right)^2$ and $\displaystyle 4x^2(x^2-a^2)$ to show that they are equal
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    10
    I also had trouble with messy algebra, but I got there in the end, using a more geometric/trigonometric approach.

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

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

    If the point (x,y) also lies on the hyperbola then $\displaystyle \frac{x^2}{a^2} - \frac{y^2}{m^2a^2} = 1$, from which $\displaystyle m^2(x^2-a^2) = y^2$. Substitute the formula for m from the previous paragraph into that, and you get the result.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Nov 2008
    Posts
    114
    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

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

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

    $\displaystyle m^4 (y^2 - x^2)^2 = 4m^2 x^2 y^2$
    $\displaystyle m^2 (y^2 - x^2)^2 = 4 x^2 y^2$

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

    $\displaystyle m^2(x^2 - y^2)^2 = 4x^2 m^2 (x^2-a^2)$
    $\displaystyle (x^2 - y^2)^2 = 4x^2(x^2 - a^2)$

    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: May 16th 2011, 04:57 AM
  2. distance of a point from a line
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: Dec 18th 2010, 11:47 PM
  3. distance of line and a point
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: Sep 4th 2009, 12:51 AM
  4. Replies: 3
    Last Post: Feb 20th 2008, 10:17 AM
  5. distance from point to line
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Nov 16th 2006, 04:30 PM

Search tags for this page

Search Tags


/mathhelpforum @mathhelpforum