finding the resutling transformed equation

[biennerienno]

Hi all

i have been working on this problem, and my answer sheet comes up with it one way, but I dont know what they did.

starting with the equation (x-h)(y-k) + K = 0. Put it in Standard Basic Form. What is the resulting transformed equation?

first off, i didnt know which set of transformations to use:

a)

y' = x cos @ + y sin @
x' = -x sin @ + y cose @

or

b)

x = x' cos @ - y' sin @
y = x' sin @ + y' cos @

@ = theta

on the answer sheet, they solve it as follows (but i do not see the in between steps and i think there may be an error somewhere like why x''y''):

let x = x'' + h, y = y'' + k.
We then have x''y'' + k = 0. Now rotate through 45 deg to eliminate x''y'' term.
This comes up with:
((x')^2 / 2) - ((y')^2 / 2) + K = 0, or ((y')^2 / 2k) - ((x')^2 / 2k) = 1


Any help you guys can give me would be great.

 

[DrPizza]

Okay,
Clearly, y-k = K/(x-h) is the hyperbola y=K/x shifted h and k units. There should be no question that theta = 45 degrees.

What they did in the first step on the answer sheet was to shift that conic section so that the center of the new conic section is on the origin... Take some time to understand how they did that. The center of the original conic section was (h,k). The center of the new conic section is now (0,0) (and hopefully we recognize it as a hyperbola, but that doesn't matter.)

The reason for this is to eliminate a LOT of extra work. (it simplifies the problem.
(I'm going to eliminate the extra marks)
So, now, you can think of the original problem as being xy - K = 0 (or xy = -K)
You're going to figure out what the equation in standard for is for that hyperbola, but you're also going to remember that later on, you'll have to shift that equation back to a center of (h,k) by replacing x with x-h and y with y-k.

Now, if you didn't shift it first, here's what you'll get:

xy -kx -hy +hk + K = 0

@=45 degrees, but again, in case you didn't realize this, for conic sections
cot2@=(A-C)/B where A is the coefficient of x^2, B is the coefficient of xy and C is the coefficient of y^2

So, Cot2@=(0-0)/1
2@=90 degrees, @=45 degrees.

If you replace the x with x'cos45 - y'sin45
which is sqrt(2)/2 * x' - sqrt(2)/2 * y'
But, I say BAHHHH to using sqrt(2)/2. Screw those math hippies... write it as x'/sqrt(2)
and replace the y with x'sin45+y'cos45
again, without a rationalized denominator, x'/sqrt(2) + y'/sqrt(2)

Now, plug them into the hard equation, and you get
(x'/sqrt(2) - y'/sqrt(2))(it's conjugate) -k(x'/sqrt(2)) +h(y'/sqrt(2)) + K = 0
Then you get
x'squared/2 - y'squared/2 -k(x'/sqrt...
Then, to get it into standard form,
you have to complete the square....

Bah, way too much work.

Shift the system to the origin,
xy+K=0
(x'/sqrt(2) - y'/sqrt(2))(it's conjugate)
x'squared/2 -y'squared/2 +K = 0

This is the system rotated. It looks to me like they forgot to shift it back though, so
(x'-h)^2/2 - (y'-h)^2/2 +K =0

I hate these problems and haven't done one in a while, so it's possible I'm overlooking something. Hopefully though, you can see why the x'' substution. I personally prefer to just change the letters... who says it has to be x and y. Then, it's easier to write.

 

[eLiu]

What level of math is this? If you've had any experience with linear algebra...there's this thign called "real quadratic forms" that makes these kinds of problems trivial.

But if you haven't...man I'm so sorry...those problems are a PITA. If it is linalg, I can help you out later on (when im not at work, lol)...otherwise, bleeegh

Edit: Whoops, I guess it really doesn't matter, since DrPizza posted the solution

 

[DrPizza]

I assumed from his equations that he was in pre-calculus. I attempted to give an answer under those conditions.

Edit: and, yeah, those problems are a PITA! I have students do 1 in class, 2 for homework, then give them another 1 and an open notes quiz on it. That's the last time they see it. A little exposure to those problems so that maybe they'll think "oh, an xy term. Those are a pain in the neck to do." Or so that at least it may ring a bell if they ever have to do it again. There are more important things to learn to do and memorize when getting ready for calculus. (I also show them an example of a degenerate case in class)

 

[eLiu]

Cool Yeah I don't know why I was thinking to linalg...I mean my comment didn't make a whole lot of sense--if he had taken linalg, he probably would've known to use it >.< Oops...

But yeah your response was a nice review for me...I think I did one of those problems in precalculus--the teacher said something like, "when and if you see this again, it will be a lot easier." Then we moved on to something else. Did you drill the unit circle, triangle relationships, and processes for obtaining the trig identities into their heads? Oh the joy of precalc! They *really* came in handy when we learned trig substitutions in calc (for integration). And they were even more useful on math contests...

But now I'm kicking off freshman year (college) with real analysis...and I'm scared :Q

Edit: What do you mean by the degenerate case? Like when the hyperbola "collapse" onto their asymptotes?

 

[JSSheridan]

Originally posted by: DrPizza
xy - K = 0 (or xy = -K)
xy -kx -hy +hk + K = 0

xy - K = 0
xy - K + K = 0 + K
xy = K;

y-k = K/(x-h)
(x-h)*(y-k) = K
xy - kx - hy + hk = K
xy - kx - hy + hk - K = 0;

It's been a long time since pre-Calc and linear though, so I don't remember how the solution goes. In fact, I don't even remember working a problem like this, but I don't really mind forgeting about it. Peace.

 

[DrPizza]

Great way to describe it, yeah, a hyperbola "collapsing" onto it's asymptotes."

I've described hyperbolas as inside out ellipses for years... it works great. Draw the ellipse, enclose it in a rectangle, draw and extend the diagonals - these are the asymptotes, then turn the ellipse inside out.

 

[biennerienno]

thanks for all the help everyone.

As for what I am taking, I am not in pre-calc. I am actually a Computer instructor who get stuck working on alot of math and electronic problems and while I understood the concepts I could not grasp the '' part of it. Then in reviewing this, I felt kinda dumb, I hate making mistakes. Thanks again!

 

[eLiu]

Originally posted by: DrPizza
Great way to describe it, yeah, a hyperbola "collapsing" onto it's asymptotes."

I've described hyperbolas as inside out ellipses for years... it works great. Draw the ellipse, enclose it in a rectangle, draw and extend the diagonals - these are the asymptotes, then turn the ellipse inside out.

Cool! My precalc teacher never showed us that...but I did notice it while doing random things with my graphing calculator.

And hrm...I've never heard a hyperbola described as an "inside out ellipse." When I go back to visit my high school teachers in the fall (before heading off to college), I'll tell them that one...I think that would relieve a lot of confusion with conic sections. Because well, everyone can think of an ellipse as a "bloated circle" but when you ask them to draw a hyperbola, it's like "wtf?" A lot of students (at least in my class) always drew two parbolas...which well, isn't right >.<

-Eric