Rotate the axes and eliminate the xy term of x^2+8xy+y^2-15=0.

a.) -3x'2+5y'2-15=0
b.) 1/5x'2-1/3y'2-15=0
c.) 5x'2-3y'2-15=0
d.) 5x'2+3y'2-15=0

Question

Rotate the axes and eliminate the xy term of x^2+8xy+y^2-15=0.
 	
a.) -3x'2+5y'2-15=0
b.) 1/5x'2-1/3y'2-15=0
c.) 5x'2-3y'2-15=0
d.) 5x'2+3y'2-15=0

tkhunny · Answer

What have you tried?  There's a cotangent in there, somewhere?

anonymous · Answer

I honestly have no idea how to do this, this an online class also so im basically on my own trying to teach myself this stuff.

tkhunny · Answer

I am becoming more and more annoyed with online instruction.  Do you have no live person at all?  Really, why should you EVER get a question for which you have "no idea"?!  It just makes no sense.

anonymous · Answer

no live person, just me.

tkhunny · Answer

Anyway, you just have to memorize these:

$x = x'\cos(\theta) - y'\sin(\theta)$
$y = x'\sin(\theta) + y'\cos(\theta)$

tkhunny · Answer

The other part is to find the angle.  It's usually found by observing that $\cot(2\theta) = \dfrac{A-C}{B}$.  This is better than using the tangent, since A = C poses quite a problem.

In your equation,  x^2+8xy+y^2-15=0, you have:

A = 1 (the coefficient on x^2)
B = 8 (the coefficient on xy)
C = 1 (the coefficient on y^2)

How are we doing, so far?

anonymous · Answer

im kind of confused, like i know what you mean by the A,B,C part but the rest is kind of confusing.

tkhunny · Answer

That's why I said you have to memorize it.  Do enough problems and you will remember better.

tkhunny · Answer

There is plenty of theory to sift through, but sometimes it's easier to do it and then think about it.  In my opinion, this is likely to be one of those times.

If $\cot(2\theta) = \dfrac{1-1}{8} = 0$, we have $2\theta = What?$  The cotangent is zero where the cosine is zero.  Please name one such angle.