# How do you use rotation of axes to identify and sketch the curve of 16x^2-8sqrt2xy+2y^2+(8sqrt2-3)x-(6sqrt2+4)y=7?

##### 1 Answer
Nov 1, 2016

$f \left(x , y\right) = 16 {x}^{2} - 8 \sqrt{2} x y + 2 {y}^{2} + \left(8 \sqrt{2} - 3\right) x - \left(6 \sqrt{2} + 4\right) y = 7$

has the structure

$\left(x , y\right) \left(\begin{matrix}{m}_{1} & {m}_{12} \\ {m}_{12} & {m}_{2}\end{matrix}\right) \left(\begin{matrix}x \\ y\end{matrix}\right) + \left({b}_{1} , {b}_{2}\right) \left(\begin{matrix}x \\ y\end{matrix}\right) + c = 0$

with

$M = \left(\begin{matrix}16 & - 4 \sqrt{2} \\ - 4 \sqrt{2} & 2\end{matrix}\right)$
$b = \left(8 \sqrt{2} - 3 , - \left(6 \sqrt{2} + 4\right)\right)$
$c = - 7$

The $M$'s characteristic polynomial qualifies the conic type.

$p \left(M\right) = {s}^{2} - 18 s = s \left(s - 18\right)$ so it is a slanted parabola because $p \left(M\right)$ has a null and a non null roots.

Calling now $p = \left(x , y\right)$ and $P = \left(X , Y\right)$ introducing a rotation such that

$P = R \cdot p$ we have

$p = {R}^{T} \cdot P$ and

$p \cdot M \cdot p + b \cdot p + c \to P \cdot R \cdot M \cdot {R}^{T} \cdot P + b \cdot {R}^{T} \cdot P + c$ or

$P \cdot {M}_{R} \cdot P + {b}_{R} \cdot P + c = 0$ is the conic in the rotated new set of coordinates. The rotation matrix $R$ is choosed such that

${M}_{R} = R \cdot M \cdot {R}^{T}$ is a diagonal matrix. ${.}^{T}$ indicates transposition which in this case is equivalent to inversion.

As we know $R = \left(\begin{matrix}\cos \theta & - \sin \theta \\ \sin \theta & \cos \theta\end{matrix}\right)$ so
M_R=((m_1 Cos^2theta - 2 m_(12) Costheta Sintheta+ m_2 Sin^2theta, m_(12) Cos(2theta) + (m_1 - m_2) Costheta Sintheta),(m_(12) Cos2theta + (m_1 - m_2) Costheta Sintheta, m_2 Cos^2theta + m_1 Sin^2theta + m_(12) Sin2theta))

Choosing now $\theta$ such that

${m}_{12} C o s 2 \theta + \left({m}_{1} - {m}_{2}\right) C o s \theta S \int h \eta = 0$ we have

$\theta = \arctan \left(\frac{2 {m}_{12}}{{m}_{1} + \sqrt{4 {m}_{12}^{2} + {\left({m}_{1} - {m}_{2}\right)}^{2}} - {m}_{2}}\right)$
we will have the rotated matrix ${M}_{R}$