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Thread: Geometric Property of Hyperbola

  1. #1
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    Exclamation Geometric Property of Hyperbola

    From a point P on x^2/a^2 - y^2/b^2 = 1, two lines are drawn parallel to each of the asymptotes, cutting the asymptotes at M and N. Prove that the area of the parallelogram ONPM is constant (where O is the origin).

    Could someone please show me how to find the two lines?
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  2. #2
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    Quote Originally Posted by xwrathbringerx View Post
    From a point P on x^2/a^2 - y^2/b^2 = 1, two lines are drawn parallel to each of the asymptotes, cutting the asymptotes at M and N. Prove that the area of the parallelogram ONPM is constant (where O is the origin).

    Could someone please show me how to find the two lines?
    The asymptotes of the hyperbola with equation $\displaystyle \frac{x^2}{a^2}- \frac{y^2}{b^2}= 1$ are given by $\displaystyle \frac{x^2}{a^2}- \frac{y^2}{b^2}= 0$.

    That factors as $\displaystyle \left(\frac{x}{a}- \frac{y}{b}\right)\left(\frac{x}{a}+ \frac{y}{b}\right)= 0$ so the two asymptotes are given by $\displaystyle y= \frac{b}{a}x$, with slope $\displaystyle \frac{b}{a}$, and $\displaystyle y= -\frac{b}{a}x$, with slope $\displaystyle -\frac{b}{a}$.

    Let P be $\displaystyle (x_0, y_0)$. The equations of the two lines through P, parallel to the asymptotes are $\displaystyle y= \frac{b}{a}(x- x_0)+ y_0$ and $\displaystyle y= -\frac{b}{a}(x- x_0)+ y_0$.

    Since P is on the hyperbola, you can use the fact that $\displaystyle \frac{x_0^2}{a^2}- \frac{y_0^2}{b^2}= 1$.
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  3. #3
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    Exclamation

    How the on earth do I do this? I found the equ. of the lines and found M and N BUT when I try to use the area of a parallelogram to prove its constant...it gets all messy and crap.

    Please show me how to do this?

    NP: bx+ay-bx1 - ay1 = 0
    MP: bx-ay-bx1+ay1=0
    N[(bx1+ay1)/2b , (bx1+ay1)/2a]
    M[(bx1-ay1)/2b,(ay1-bx1)/2a]
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  4. #4
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    Hmmm I get - 1 / 2ab for the constant for some reason ... even though it's meant to be an area ...
    Last edited by Sunyata; Mar 5th 2010 at 04:55 PM.
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  5. #5
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    Hello, xwrathbringerx!

    Your work is correct . . .

    I used: .$\displaystyle P(h,k)$
    (Subscripts always confuse me.)

    So I have: .$\displaystyle \begin{Bmatrix}O:& (0,0) \\ \\[-3mm] P: & (h,k) \\ \\[-3mm] M: & \left(\dfrac{bh+ak}{2b},\:\dfrac{bh+ak}{2a}\right) \end{Bmatrix}$

    Then I used vectors: .$\displaystyle \begin{Bmatrix} \overrightarrow{OP}: & \langle h,k\rangle \\ \\[-3mm]\overrightarrow{OM}:& \left\langle \dfrac{bh+ak}{2b},\:\dfrac{bh+ak}{2x}\right\rangle \end{Bmatrix} $


    Parallelgrom $\displaystyle ONPM$ is comprised of two congrent triangles:
    . . $\displaystyle \Delta OMP$ and $\displaystyle \Delta ONP.$


    Formula: . $\displaystyle \text{Area }\Delta OPM \;=\;\tfrac{1}{2}\left|\overrightarrow{OP} \times \overrightarrow{OM }\right| $


    . . . . . . . . $\displaystyle =\;\frac{1}{2}\left|\begin{array}{cc}\dfrac{bh+ak} {2b} & \dfrac{bh+ak}{2a} \\ \\[-3mm] h & k \end{array}\right| \;=\;\frac{bh+ak}{4}\left|\begin{array}{cc}\dfrac{ 1}{b} & \dfrac{1}{a} \\ \\[-3mm] h & k \end{array}\right|$


    . . . . . . . . $\displaystyle =\;\frac{bh+ak}{4}\left(\frac{k}{b} - \frac{h}{a}\right) \;=\;\frac{bh+ak}{4}\left(\frac{bh+ak}{ab}\right)$


    . . . . . . . . $\displaystyle =\;\frac{b^2h^2 - a^2k^2}{4ab}$

    Hence, Area of parallelogram $\displaystyle ONPM\!:\;\;A \;=\;2\times \frac{b^2h^2 - a^2k^2}{4ab} \;=\;\frac{b^2h^2 - a^2k^2}{2ab}$ .[1]



    Point $\displaystyle (h,k)$ is on the hyperbola .$\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} \:=\:1$

    . . Hence: .$\displaystyle \frac{h^2}{a^2} - \frac{k^2}{b^2} \:=\:1 \quad\Rightarrow\quad b^2h^2 - a^2k^2 \:=\:a^2b^2$ .[2]


    Substitute [2] into [1]: .$\displaystyle A \;=\;\frac{a^2b^2}{2ab} \;=\;\tfrac{1}{2}ab$ . . . a constant!

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  6. #6
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    Here's another way...

    The parallelogram area is

    (Area of triangle oqc)-(Area of triangle spc)

    To find c

    $\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=1\ \Rightarrow\ \frac{x^2}{a^2}=1+\frac{y^2}{b^2}=\frac{b^2+y^2}{b ^2}$

    $\displaystyle \Rightarrow\ x^2=\frac{a^2\left(b^2+y^2\right)}{b^2}$

    When x=c, y=p on the hyperbola

    $\displaystyle c=\frac{a}{b}\sqrt{b^2+p^2}$

    q

    When x=c, y=q on the asymptote

    $\displaystyle q=\frac{b}{a}c=\sqrt{b^2+p^2}$

    s

    Equation of the parallel line from "p" with positive slope

    $\displaystyle y-p=\frac{b}{a}\left(x-\frac{a}{b}\sqrt{b^2+p^2}\right)$

    y=0, x=s

    $\displaystyle -p=\frac{b}{a}s-\sqrt{b^2+p^2}\ \Rightarrow\ \sqrt{b^2+p^2}-p=\frac{b}{a}s\ \Rightarrow\ s=\sqrt{b^2+p^2}-\frac{a}{b}p$

    Parallelogram area

    $\displaystyle \frac{1}{2}|cq|-\frac{1}{2}|c-s|p=\frac{1}{2}\left(\frac{a}{b}\sqrt{b^2+p^2}\rig ht)\sqrt{b^2+p^2}-\frac{1}{2}\left(\frac{a}{b}\sqrt{b^2+p^2}-\frac{a}{b}\sqrt{b^2+p^2}+\frac{a}{b}p\right)p$

    $\displaystyle =\frac{1}{2}\frac{a}{b}\left(a^2+b^2\right)-\frac{1}{2}\frac{a}{b}p^2=\frac{1}{2}\frac{a}{b}\l eft(b^2+p^2-p^2\right)$

    $\displaystyle =\frac{1}{2}ab=constant$
    Attached Thumbnails Attached Thumbnails Geometric Property of Hyperbola-asymptotes-hyperbola.jpg  
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