# Thread: Geometric Property of Hyperbola

1. ## 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?

2. Originally Posted by xwrathbringerx
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 $\frac{x^2}{a^2}- \frac{y^2}{b^2}= 1$ are given by $\frac{x^2}{a^2}- \frac{y^2}{b^2}= 0$.

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

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

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

3. 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]

4. Hmmm I get - 1 / 2ab for the constant for some reason ... even though it's meant to be an area ...

5. Hello, xwrathbringerx!

Your work is correct . . .

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

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

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

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

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

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

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

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

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

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

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

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

6. Here's another way...

The parallelogram area is

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

To find c

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

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

When x=c, y=p on the hyperbola

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

q

When x=c, y=q on the asymptote

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

s

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

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

y=0, x=s

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

Parallelogram area

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

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

$=\frac{1}{2}ab=constant$