Question

To transform the equation

$A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0,B\ne 0$

into one in $x\text{'}$and $y\text{'}$without an $x\text{'}y\text{'}-$term, rotate the axes

through an acute angle $\theta$that satisfies the equation _______.

Short answer

Answer is$\cos \left(2\theta \right)=\frac{A-C}{B}$.

Step-by-step solution

Step 1. Given information

An equation is$A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0,B\ne 0$.

Step 2. Fill in the blank.

The rotation of axes through an appropriate angle $\theta$will transform the equation $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$into an equation in $x\text{'}$and $y^{\text{'}}$without an $x^{2}y$term.

In order to find the formula for choosing an appropriate angle $\theta$, we use the rotation formulas in the above equation.

$A{\left(x\text{'}\cos \theta -y\text{'}\sin \theta \right)}^2+B\left(x\text{'}\cos \theta -y\text{'}\sin \theta \right)\left(x\text{'}\sin \theta +y\text{'}\cos \theta \right)+C{\left(x\text{'}\sin \theta +y\text{'}\cos \theta \right)}^2+D\left(x\text{'}\cos \theta -y\text{'}\sin \theta \right)+E\left(x\text{'}\sin \theta +y\text{'}\cos \theta \right)+F=0$

Expand and collecting the like terms.$\left(A{\cos }^2\theta +B\sin \theta \cos \theta +C{\sin }^2\theta \right){\left(x\text{'}\right)}^2+\left[B\left({\cos }^2\theta -{\sin }^2\theta +2\left(C-A\right)\sin \theta \cos \theta \right)\right]x\text{'}y\text{'}+\left(A{\sin }^2\theta -B\sin \theta \cos \theta +C{\cos }^2\theta \right){\left(y\text{'}\right)}^2+\left(D\cos \theta +E\sin \theta \right)x\text{'}+\left(-D\sin \theta +E\cos \theta \right)y\text{'}+F=0$Since we want to eliminate the $x^{2}y$term from this equation, we select an angle $\theta$, so that the coefficient of the $x^{2}y$term, $B\left({\cos }^2\theta -{\sin }^2\theta \right)+2(C-A)\sin \theta \cos \theta$is $0$.

$\begin{array}{r}B\left({\cos }^2\theta -{\sin }^2\theta \right)+2(C-A)\sin \theta \cos \theta =0\\ B\cos \left(2\theta \right)+(C-A)\sin \left(2\theta \right)=0\end{array}$

role="math" localid="1646661431364" $\begin{array}{rcl}B\cos \left(2\theta \right)& =& (A-C)\sin \left(2\theta \right)\\ \cot \left(2\theta \right)& =& \frac{A-C}{B};B\ne 0\end{array}$

Therefore, to transform the equation $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0,B\ne 0$into one and$y\text{'}$ without an$x^{2}y$term, rotate the axes through an acute angle $\theta$that satisfies the equation $\cot \left(2\theta \right)=\frac{A-C}{B}$