Welcome to our community

Be a part of something great, join today!

Find rs: xy = 1 reflected across y = 2x

  • Thread starter
  • Admin
  • #1

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,755
Hi MHB,

I recently found an interesting problem which I initially thought I could solve without much trouble, but after I started to work on it, I found out that this has not been the case and this problem has driven me crazy, as I find no good way to attack it...so, I desperately need some help on this problem.

Problem:

The graph $xy=1$ is reflected in $y=2x$ to give the graph of $12x^2+rxy+sy^2+t=0$. Find $rs$.

I tried to sketch its graph and here is my sketch:

Find rs.JPG

I managed only to get down to finding the relation between $r$ and $s$, i.e. $(3r+s=-33)$, but not the value of the product of them.

Thanks in advance for any help that anyone can offer.







 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Find rs

Have you considered that the reflection matrix is:

\(\displaystyle R(m)=\frac{1}{5}\begin{bmatrix}{-3}&{4}\\{4}&{3}\end{bmatrix}\)
 
  • Thread starter
  • Admin
  • #3

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,755
Re: Find rs

Have you considered that the reflection matrix is:

\(\displaystyle R(m)=\frac{1}{5}\begin{bmatrix}{-3}&{4}\\{4}&{3}\end{bmatrix}\)
Hi MarkFL,

Thanks for the reply but unfortunately, I'm not familiar with matrices...
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Find rs

Okay, what we want to compute, is the product:

\(\displaystyle \frac{1}{5}\begin{bmatrix}{-3}&{4}\\{4}&{3}\end{bmatrix}\begin{bmatrix}{x}\\{y}\end{bmatrix}=\frac{1}{5}\begin{bmatrix}{-3x+4y}\\{4x+3y}\end{bmatrix}\)

And so $xy=1$ becomes:

\(\displaystyle \left(\frac{1}{5}(-3x+4y) \right)\left(\frac{1}{5}(4x+3y) \right)=1\)

Can you proceed from here? (Sun)
 
  • Thread starter
  • Admin
  • #5

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,755
Re: Find rs

Okay, what we want to compute, is the product:

\(\displaystyle \frac{1}{5}\begin{bmatrix}{-3}&{4}\\{4}&{3}\end{bmatrix}\begin{bmatrix}{x}\\{y}\end{bmatrix}=\frac{1}{5}\begin{bmatrix}{-3x+4y}\\{4x+3y}\end{bmatrix}\)

And so $xy=1$ becomes:

\(\displaystyle \left(\frac{1}{5}(-3x+4y) \right)\left(\frac{1}{5}(4x+3y) \right)=1\)

Can you proceed from here? (Sun)
I'll try my best, hehehe...

If we expand and cross multiply the equation below, we get

\(\displaystyle \left(\frac{1}{5}(-3x+4y) \right)\left(\frac{1}{5}(4x+3y) \right)=1\)

\(\displaystyle (-3x+4y)(4x+3y)=25\)

\(\displaystyle -12x^2-9xy+16xy+12y^2=25\)

\(\displaystyle 12x^2-7xy-12y^2+25=0\)

and now, by comparing it with the given equation as follows:

$12x^2+rxy+sy^2+t=0$

We see that $r=-7$ and $s=-12$, and therefore $rs=(-7)(-12)=84$.

But when substituting these values to the equation that relates $r$ and $s$ gives

$3r+s=3(-12)+(-7)=-43\ne -33$, why is that so? Was it because the equation that I found wasn't correct?:confused: But...I checked and I'm very certain that I got the right equation of $3r+s=-33$...
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Find rs

I haven't checked your work thoroughly, but one error I see is:

\(\displaystyle \tan\left(\pi-2\tan^{-1}(2) \right)\ne\frac{2}{11}\)

This should be:

\(\displaystyle \tan\left(\pi-2\tan^{-1}(2) \right)=\frac{4}{3}\)

But, this still does not resolve the issue.
 
  • Thread starter
  • Admin
  • #7

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,755
Re: Find rs

Ops...my bad...I labeled the diagram, the angle between the gradient line of the reflected graph at the intersection point wrongly...:eek:

I've uploaded a corrected version of the graph, as shown below:
Find rs (corrected).JPG
and now, after substituting the values of $r=-7$ and $s=-12$ into the equation

\(\displaystyle \frac{dy}{dx}=\frac{-2r-24}{r+4s}\)

we get

\(\displaystyle \frac{dy}{dx}=\frac{-10}{-55}=\frac{2}{11}\)

but when I tried it with the equation that I simplified, which is $3r+s=-33$, I got $3(-7)+(-12)=-43$...

And I don't know what went wrong with the simplified equation that I obtained...(Angry)
 
  • Thread starter
  • Admin
  • #8

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,755
Re: Find rs

Okay, what we want to compute, is the product:

\(\displaystyle \frac{1}{5}\begin{bmatrix}{-3}&{4}\\{4}&{3}\end{bmatrix}\begin{bmatrix}{x}\\{y}\end{bmatrix}=\frac{1}{5}\begin{bmatrix}{-3x+4y}\\{4x+3y}\end{bmatrix}\)

And so $xy=1$ becomes:

\(\displaystyle \left(\frac{1}{5}(-3x+4y) \right)\left(\frac{1}{5}(4x+3y) \right)=1\)

Can you proceed from here? (Sun)
Hi MarkFL,

Sorry to bother you again...as I don't know the concept behind all these...would you mind to explain it for me?

Where does the term \(\displaystyle \frac{1}{5}\) and \(\displaystyle \frac{1}{5}\begin{bmatrix}{-3}&{4}\\{4}&{3}\end{bmatrix}\) come from?:confused:
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Find rs

But \(\displaystyle 3(-7)+(-12)=-21-12=-33\)...(Clapping)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,755
Re: Find rs

But \(\displaystyle 3(-7)+(-12)=-21-12=-33\)...(Clapping)
Argh!:mad: I couldn't believe I made such a silly mistake here...and thanks for catching this up for me!

Here is the article I referenced:

http://www.faculty.sfasu.edu/becneljj/homepage/preprints/reflection.pdf

Look particularly at pages 5 and 6.
Hey, thanks for the link! I really appreciate that! And thanks again for the help!


Hey MarkFL, yesterday, when I asked you privately if you have any idea on how to tackle this problem, you said you were not interested to it...but now you're the one who replied to me... I hate you! (Tongueout) Now I want you to make some crunchy popcorn and white coffee for me!(Coffee)(Emo)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Find rs

...
Hey MarkFL, yesterday, when I asked you privately if you have any idea on how to tackle this problem, you said you were not interested to it...but now you're the one who replied to me... I hate you! (Tongueout) Now I want you to make some crunchy popcorn and white coffee for me!(Coffee)(Emo)
Reviewing our conversation, I see that I actually did not reply ( (Tongueout) ) because I was doing other things at that time, and then I quickly forgot about it, until you told me you were going to post it here. So, at that time, having completed my "duties," I began to research the technique of reflections about the line $y=mx$, and found that helpful PDF, from which I was able to figure out what could be done to find the answer to the given problem.

In the end though, I found this problem to be interesting, and I am glad you wound up posting it here because others may benefit from it. So, I will make that popcorn and coffee for you, but I will need you to bring me some refreshing Dew instead for fudging about me saying I was not interested. (Sun)
 

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,755
Re: Find rs

Hehehe...alright, I accept your reasoning and I'll bring you the watered down Dew instead...:p(Emo)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Watered down?!? (Puke) (Swearing) Remind me to keep my eye on you...(Tongueout)

By the way, I suspect there may be another linear equation in $(r,s)$ lurking around in your method.

I am also wondering if there is a more clever technique than either of us is using which gives us the desired product without having to find the factors explicitly...(Thinking)

Fun problem! (Yes)
 

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,755
Hey our naughty and playful global moderator, aka MarkFL,

You're right, I found another way to solve for the values for $r$ and $s$ using my method.

My solution:
Find rs_2.JPG

The line $y=2x$ is the perpendicular bisector of the segment AB and this fact leads us to many useful results:

1)

The gradient between the point $A(a, \frac{1}{a})$ and $(1, 2)$ tells us the value for $a$...

\(\displaystyle \frac{2-\frac{1}{a}}{1-a}=-\frac{1}{2}\) implies \(\displaystyle a=\frac{5-\sqrt{17}}{2}\).

and

The gradient between the point $B(m, k)$ and $(1, 2)$ tells us the relation between $m$ and $k$...

\(\displaystyle \frac{k-2}{m-1}=-\frac{1}{2}\) implies \(\displaystyle m=5-2k\).


2)

We see that the curve of $12x^2+rxy+sy^2+t=0$ passes through the intersection point\(\displaystyle \left(\frac{1}{\sqrt{2}},\;\;\frac{2}{\sqrt{2}} \right)\) between the curve $xy=1$ and the line $y=2x$ and if we merge all these data together with the relation between $r$ and $s$ where $3r+s=-33$, we get

$t=5r+60$

3)

The exact same distance between the point \(\displaystyle A(a, \frac{1}{a})=\left(\frac{5-\sqrt{17}}{2},\; \frac{2}{5-\sqrt{17}} \right)\), $(1, 2)$ and $B(m, k)=(5-2k, k)$, $(1, 2)$ tells us the value for k...

\(\displaystyle \left(2-\frac{2}{5-\sqrt{17}} \right)^2+\left(1-\frac{5-\sqrt{17}}{2} \right)^2=(k-2)^2+(5-2k-1)^2\)

Solving the equation above we get \(\displaystyle k=\frac{11-\sqrt{17}}{4}\)

At this point, we have all that we need to solve for the value of $r$.

\(\displaystyle 12\left(\frac{\sqrt{17}-1}{2} \right)^2+r\left(\frac{\sqrt{17}-1}{2} \right)\left(\frac{11-\sqrt{17}}{4} \right)-(33+3r)\left(\frac{11-\sqrt{17}-1}{4} \right)^2+5r+60=0\)

$r=-7$

$s=-33-3(-7)=-12$

and hence

$rs=(-7)(-12)=84$ and we're done!
 
Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
I am also wondering if there is a more clever technique than either of us is using which gives us the desired product without having to find the factors explicitly...(Thinking)
I just wanted to add that the condition for the general conic equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ to represent a rectangular hyperbola is $a+b=0$. In this case, that tells you immediately that $s = -12$. But I do not see any quick way to find the value of $r$, or to get $rs$ without finding $r$ and $s$ separately.
 

johng

Well-known member
MHB Math Helper
Jan 25, 2013
236
Hi,
Here's my version of reflection about a line:

MHBgeometry1.png
 

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,755