# Help! Rotate the axis to eliminate the xy-term in the equation then write the equation ins standard form. x^2-4xy+4y^2+5sqrt(5)+1=0 Help!?

Dec 3, 2016

#### Explanation:

The equation given here ${x}^{2} - 4 x y + 4 {y}^{2} + 5 \sqrt{5} + 1 = 0$ is of the form $A {x}^{2} + B x y + C {y}^{2} + D x + E y + F = 0$.

What a rotation does is it changes $x$ & $y$-axes to $x '$ & $y '$-axes, as shown below,.

In such a case, the relation between coordinate $\left(x , y\right)$ and new coordinates $\left(x ' , y '\right)$ is given by

$x = x ' \cos \theta - y ' \sin \theta$ and $y = x ' \sin \theta + y ' \cos \theta$

and reverse is $x ' = x \cos \theta + y \sin \theta$ and $y ' = - x \sin \theta - \cos \theta$

Note that latter equations are equivalent to rotation by $- \theta$.

In such cases, we can eliminate $x y$ if rotated by $\theta = \frac{A - C}{B}$

In given equation we have $\cot 2 \theta = \frac{1 - 4}{- 4} = \frac{3}{4}$ i.e.

$\frac{{\cot}^{2} \theta - 1}{2 \cot \theta} = \frac{3}{4}$ or $4 {\cot}^{2} \theta - 6 \cot \theta - 4 = 0$

or $\left(2 \cot \theta - 4\right) \left(2 \cot \theta + 1\right) = 0$ i.e. $\cot \theta = 2$ or $- \frac{1}{2}$

These two angles relate to $\theta$ and $\theta - {90}^{o}$ in the image above. Working out for $\cot \theta = 2$

Hence, either $\sin \theta = \frac{1}{\sqrt{5}}$ and $\cos \theta = \frac{2}{\sqrt{5}}$

or $\sin \theta = - \frac{2}{\sqrt{5}}$ and $\cos \theta = \frac{1}{\sqrt{5}}$

and we have $x = \frac{2 x '}{\sqrt{5}} + \frac{y '}{\sqrt{5}}$ and $y = \frac{x '}{\sqrt{5}} + \frac{2 y '}{\sqrt{5}}$

and putting these in given equation and simplifying we get

$9 {y}^{2} + 25 \sqrt{5} + 5 = 0$

One can also try for $\cot \theta = - \frac{1}{2}$, for which we get $\sin \theta = - \frac{2}{\sqrt{5}}$ and $\cos \theta = \frac{1}{\sqrt{5}}$

$x ' = x \cos \theta + y \sin \theta$ and $y ' = - x \sin \theta - \cos \theta$

i.e $x ' = \frac{x}{\sqrt{5}} - \frac{2 y}{\sqrt{5}}$ and $y ' = \frac{2 x}{\sqrt{5}} - \frac{y}{\sqrt{5}}$

and simplifying $9 {x}^{2} + 25 \sqrt{5} = 5 = 0$

Note - Please observe that above equation ${x}^{2} - 4 x y + 4 {y}^{2} + 5 \sqrt{5} + 1 = 0$

$\Leftrightarrow {\left(x - 2 y\right)}^{2} + + 5 \sqrt{5} + 1 = 0$ and as LHS for $x \in \mathbb{R}$ and $y \in \mathbb{R}$ is always positive, does not have real solution and as such cannot be represented on Cartesian Plane.

Dec 3, 2016

See below.

#### Explanation:

This possible conic can be written as

$C \to \left(x , y\right) \left(\begin{matrix}1 & - 2 \\ - 2 & 4\end{matrix}\right) \left(\begin{matrix}x \\ y\end{matrix}\right) + 5 \sqrt{5} + 1 = 0$

Making a coordinate change such that

$\left(\begin{matrix}X \\ Y\end{matrix}\right) = R \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}\cos \theta & - \sin \theta \\ \sin \theta & \cos \theta\end{matrix}\right) \left(\begin{matrix}x \\ y\end{matrix}\right)$ we get

${C}_{R} \to \left(X , Y\right) R . M . {R}^{T} \left(\begin{matrix}X \\ Y\end{matrix}\right) + 5 \sqrt{5} + 1 = 0$ or

${C}_{R} \to \left(X , Y\right) {M}_{R} \left(\begin{matrix}X \\ Y\end{matrix}\right) + 5 \sqrt{5} + 1 = 0$ with

${M}_{R} = R . M . {R}^{T}$ resulting in

${C}_{R} \to \left(\frac{5}{2} - \frac{3}{2} \cos \left(2 \theta\right) + 2 \sin \left(2 \theta\right)\right) {X}^{2} - \left(4 \cos \left(2 \theta\right) + 3 \sin \left(2 \theta\right)\right) X Y + \left(\frac{5}{2} + \frac{3}{2} \cos \left(2 \theta\right) - 2 \sin \left(2 \theta\right)\right) {Y}^{2} + 5 \sqrt{5} + 1 = 0$

Choosing $\theta$ such that the coefficient of $X Y$ is null or

$4 \cos \left(2 \theta\right) + 3 \sin \left(2 \theta\right) = 0$ we get

$\theta = - \frac{1}{2} \arctan \left(4 , 3\right)$

Now substituting this value into the rotation we have

${C}_{R} \to 5 {Y}^{2} + 5 + \sqrt{5} + 1 = 0$

This final result shows that ${C}_{R}$ is not a real conic because there is not real solution satisfying it.

Dec 3, 2016

Please check the original equation. I think there is something wrong.

#### Explanation:

Here is a reference on Rotation of Axes

The general form of a conic is:

$A {x}^{2} + B x y + C {y}^{2} + D x + E y + F = 0$

For the given equation:

$A = 1$
$B = - 4$
$C = 4$
$D = 0$
$E = 0$
$F = 5 \sqrt{5} + 1$

The angle of rotation is:

$\theta = \frac{1}{2} {\tan}^{-} 1 \left(\frac{B}{C - A}\right)$

$\theta = \frac{1}{2} {\tan}^{-} 1 \left(- \frac{4}{4 - 1}\right)$

$\theta = \frac{1}{2} {\tan}^{-} 1 \left(- \frac{4}{3}\right)$

$\theta \approx - {26.565}^{\circ}$

Using equations from the reference:

A' = (A + C)/2 + [(A - C)/2] cos(2θ) - B/2 sin(2θ)
$A ' = \frac{1 + 4}{2} + \left[\frac{1 - 4}{2}\right] \cos \left(- 53.13\right) - - \frac{4}{2} \sin \left(- 53.13\right)$
$A ' = \frac{5}{2} - \frac{3}{2} \cos \left(- 53.13\right) + 2 \sin \left(- 53.13\right)$
$A ' = 0$
$B ' = 0$
C' = (A + C)/2 + [(C - A)/2] cos 2θ + B/2 sin 2θ
$C ' = 3.2$
$D ' = 0$
$E ' = 0$
$F ' = F = 5 \sqrt{5} + 1$

I think there is something wrong; $A ' = 0$ should not be.
When I try to graph either the original equation or the rotated equation, using Desmos.com, I get nothing.