Eliminating the xy-Term: Solving Rotating Axes Problem with θ = 30 degrees

[orcbum]

"Rotate the axes to eliminate the xy-term in the equation. Then write the equation in standard form. (Use xp for x and yp for y' in your answer. Rotate the coordinate axes through an angle θ with 0 ≤ θ ≤ π/2.)"

13x² + 6√3xy + 7y² – 32 = 0

I figured out that θ is 30 degrees. After simplifying I still could not eliminate the xy-term: I ended up with:

10x² - 6xy + 10y² + 3(sqrt(3))x² - 3(sqrt(3))y² -32 = 0

note: the x and y terms above are in prime form (I just don't know how to show that on a forum)

As you can see I still have a negative 6xy. Considering the previous HW questions I've solved eliminated the xy terms easily, I'm at a loss with what to do should I encounter leftover xy terms.

Thanks in advance.

EDIT: You know this always happens to me, whenever I just ask a question/post one, I've solved it. I'm sorry, but I've figured it out now. (Calculating an earlier step incorrectly).

However, a question still stands. Provided the equation doesn't solve out neatly and has left over xy terms, is there a way around it?

 

[MarkFL]

...
However, a question still stands. Provided the equation doesn't solve out neatly and has left over xy terms, is there a way around it?

An appropriate rotation of axes enables us to eliminate the $xy$-term in:

(1) \(\displaystyle ax^2+bxy+cy^2+dx+ey+f=0\)

and obtain an equation of the form:

(2) \(\displaystyle AX^2+CY^2+DX+EY+F=0\)

If (2) defines a real locus of points other than a point, a line, or a pair of intersecting lines, its graph will then be a conic section. The cases: a point, a line, and a pair of intersecting lines are said to be degenerate conics.

Consider:

\(\displaystyle x=C\cos(\theta)-Y\sin(\theta)\)

\(\displaystyle y=X\sin(\theta)+Y\cos(\theta)\)

Substituting these equations into (1) and simplifying reveal that the coefficient of the $XY$-term are zero provided that:

\(\displaystyle 2(c-a)\sin(\theta)\cos(\theta)+b\left(\cos^2(\theta)-\sin^2(\theta) \right)=0\)

or equivalently:

(3) \(\displaystyle (a-c)\sin(2\theta)=b\cos(2\theta)\)

Notice that if $a=c$ in (1), then $a-c=0$ in (3). The resulting equation, $\cos(2\theta)=0$ implies that a rotation angle of $\theta=\dfrac{\pi}{4}$ will eliminate the $xy$-term in (1).

However, if $a\ne c$, then this elimination can be accomplished by choosing $\theta$ to be an angle for which:

\(\displaystyle \tan(2\theta)=\frac{b}{a-c}\)

 

[topsquark]

EDIT: You know this always happens to me, whenever I just ask a question/post one, I've solved it. I'm sorry, but I've figured it out now. (Calculating an earlier step incorrectly).

However, a question still stands. Provided the equation doesn't solve out neatly and has left over xy terms, is there a way around it?

There are some derivations that come out "smoothly." The one in your example here is what I call "fussy": it is easy to make a mistake and can be very hard to find. If you do the job correctly there will not be an xy term. If you have one left over then you have made a mistake somewhere.

The only advice I can give here is to write out every single step. All of them. One step, one term at a time. It's a pain in the pah-tootie but you can find any mistakes easier this way.

-Dan