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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Mao on May 21, 2010, 12:50:08 pm

Title: A challenge (kinda)
Post by: Mao on May 21, 2010, 12:50:08 pm: To current year 11/12 kids, this problem was quite interesting:

There are three arbitrary points P, Q and R in $\mathbb{R}^2$ space. Points P and Q are on the x axis equidistant from the origin. The x coordinate of point R lies between P and Q, the y coordinate of R is unspecified. It is always possible to construct an ellipse through these three points.

Let $\vec{r}_{PQ}$ be the displacement vector from P to Q, and similarly define other displacement vectors. We can show that the ellipse will always be in the form of

$x^2 + \frac{1}{C}y^2 = \left( \frac{1}{2} |\vec{r}_{PQ}| \right)^2$ , where C is some constant depending on the position of R.

Find an expression for C in terms of the displacement vectors (do not use scalars, i.e. do not say "let $|\vec{r}_{PQ}| = a$ , ....").

:)
Title: Re: A challenge (kinda)
Post by: theuncle on May 22, 2010, 11:09:20 am: Ive done the working in my book but i thought i'd try the answer first incase it's wrong! (r is the position vector of R)
does $C=\frac{|r|^2}{(\frac{1}{2}|\vec{PQ}|)^2}$
Title: Re: A challenge (kinda)
Post by: kyzoo on May 22, 2010, 05:13:17 pm: I think your answer is correct only if R is on the y-axis.
Title: Re: A challenge (kinda)
Post by: theuncle on May 22, 2010, 06:06:03 pm: Ok, so i'll explain how i got it...
we know that
1- $x^2+y^2=|r|^2$
also, 2- $x^2 + \frac{1}{C}y^2 = \left( \frac{1}{2} |\vec{r}_{PQ}| \right)^2$
from here it is a matter of solving simultaneous equations by subbing in known points of x and y
we know that when $x=0, |y|=|r|, y^2=|r|^2$

subbing this into equation 2, $\frac{r^2}{C} = \left( \frac{1}{2} |\vec{r}_{PQ}| \right)^2$
therefore...
$C=\frac{|r|^2}{(\frac{1}{2}|\vec{r}_{PQ}|)^2}$
this also holds for y=0
as far as I can see... I've given an equation of an ellipse with (0,0) centre that gives correct values when x=0 and y=0,
I figured no other equation could give an ellipse that had the same max x and y values... but it wouldn't be the first time i'm wrong ;)
Title: Re: A challenge (kinda)
Post by: kyzoo on May 22, 2010, 08:55:18 pm: Your first equation describes a circle, but R, P, and Q form an ellipse
Title: Re: A challenge (kinda)
Post by: theuncle on May 22, 2010, 09:31:09 pm: It would be a circle if r was a constant, but because r changes, ie. the radius of the circle changes... the equation i gave was infact that of an ellipse.

...C is constant, but when we fix r at a point,
$x^2+y^2=|r|^2$ holds for that point. And that's all that matters
Title: Re: A challenge (kinda)
Post by: Mao on May 22, 2010, 10:53:52 pm: I have no idea if my expression will simplifies to yours when R lies on the y axis. But regardless, that assumption cannot be made. The point R is any point lying between the x coordinates of P and Q with an arbitrary y coordinate.
Title: Re: A challenge (kinda)
Post by: googoo on May 22, 2010, 11:52:36 pm: Let R be vector OR, r be vector OQ and u be unit vector parallel to vector OQ.

C = (|R|^2 - (R.u)^2) / (|r|^2 - (R.u)^2)
Title: Re: A challenge (kinda)
Post by: Mao on May 23, 2010, 09:31:00 pm: Displacement vectors? :) You've given the answer in terms of position vectors.
Title: Re: A challenge (kinda)
Post by: googoo on May 23, 2010, 10:52:29 pm: R = vectorPR - r, r = 0.5vectorPQ

By the way, R and r are not position vectors. I considered them as displacement vectors from O to R and Q respectively.