# Solve and Graph 13x^2+ 6√3xy + 7y^2 – 16 = 0

• March 22nd 2011, 02:35 PM
outlawbtmn
Solve and Graph 13x^2+ 6√3xy + 7y^2 – 16 = 0
13x^2+ 6√3xy + 7y^2 – 16 = 0

I'm sorry if this isn't algebra, I assumed it was. If it is Calculus or something else, please direct me to the right place.
I'm looking for the answer, but I'm looking for how to do it more than I am searching for the answer. So please be thorough in your response.
• March 22nd 2011, 02:37 PM
e^(i*pi)
Rearrange into $7y^2 + 6\sqrt{3}xy + 13x^2 - 16 = 0$. Now it's in the form $ax^2+bx+c=0$ you can solve for y in terms of x
• March 22nd 2011, 02:41 PM
outlawbtmn
y(6√3x +7y) + 13x^2 – 16 = 0

am I getting closer?
• March 22nd 2011, 02:57 PM
topsquark
Quote:

Originally Posted by outlawbtmn
y(6√3x +7y) + 13x^2 – 16 = 0

am I getting closer?

I'm pretty sure he meant using the quadratic formula. a = 7, b = 6*sqrt(3), c = 13x^2 - 16

However, I note that this is an ellipse. Instead of "solve" did you mean simplify?

-Dan
• March 22nd 2011, 03:02 PM
outlawbtmn
Wow, okay I feel dumb now. When I said solve, I meant help me find the information I need in order to graph it.
Thanks everyone!
• March 22nd 2011, 03:16 PM
TheEmptySet
Quote:

Originally Posted by outlawbtmn
13x^2+ 6√3xy + 7y^2 – 16 = 0

I'm sorry if this isn't algebra, I assumed it was. If it is Calculus or something else, please direct me to the right place.
I'm looking for the answer, but I'm looking for how to do it more than I am searching for the answer. So please be thorough in your response.

Since you put this in the Linear Algebra section I am assuming you know what a quadratic form is. This can be rewritten in matrix form as

$\begin{bmatrix}x & y \end{bmatrix}\begin{bmatrix}13 & 3\sqrt{3} \\ 3\sqrt{3} & 7 \end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix}=16$

Now the coefficient matrix is real symmetric so the matrix is orthogonally diagonalizable.

Now diagonalize the coefficient matrix I got the two eigenvalues

$\lambda =4,16$

Find the two linearly independent eigenvectors and normalize them to have length 1.

This will give you the orthogonal matrix

$P=\frac{1}{2}\begin{bmatrix}1& \sqrt{3} \\ -\sqrt{3} & 1\end{bmatrix}$

Now back to the equation.

$\begin{bmatrix}x & y \end{bmatrix}\begin{bmatrix}13 & 3\sqrt{3} \\ 3\sqrt{3} & 7 \end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix}=16$

written in matrix form gives

$\mathbf{x}^TA\mathbf{x}=16$

Remember that for an orthogonal matrix its transpose is its inverse now make the substitution

$\mathbf{x}=P\mathbf{z} \implies \mathbf{x}^T=(P\mathbf{z})^T=\mathbf{z}^TP^T=\math bf{z}^TP^{-1}$

subbing this into the above equation gives

$\mathbf{z}^TP^{-1}AP\mathbf{z}=16$

Which now gives a diagonal matrix in our new coordinate system we have

$\begin{bmatrix}x' & y' \end{bmatrix}\begin{bmatrix}4 & 0 \\ 0 & 16 \end{bmatrix}\begin{bmatrix} x' \\ y'\end{bmatrix}=16$

OR

$4(x')^2+16(y')^2=16 \iff \frac{(x')^2}{4}+(y')^2=1$ which is an ellipse

$P=\frac{1}{2}\begin{bmatrix}1& \sqrt{3} \\ -\sqrt{3} & 1\end{bmatrix}=\begin{bmatrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta)\end{bmatrix}$

This gives $\displaystyle \theta=\frac{\pi}{3}$

This can now be plotted in the rotated coordinate system.