1. ## Rotation of Conic

Hi,

Problem: Rotate the axes so that the new equation contains no xy-term. Discuss the new equation.

x^2 + 2xy + y^2 - 8x + 8y = 0

Here is what I got so far:

Solution:

A = 1, B = 2, C =1

cot 2 theta = (A - C) / B ==> (1 - 1)/2 ==> 0/2
(cot^2 theta - 1)/ 2 cot theta = 0/2
2 cot^2 theta - 2 = 0
2(cot^2 theta - 1) = 0
2(cot theta - 1) (cot theta + 1) = 0
cot theta = 1
theta = 45 degree

x = x' cos theta - y' sin theta
x = x' cos 45 - y' sin 45

y = x' sin 45 + y' cos 45

i got to this far and plug in to the original equation and simplied it. But I don't know what to do after that. Please help me!

Thank you!

2. Hello, fallingsky!

Rotate the axes so that the new equation contains no xy-term.
. . $
x^2 + 2xy + y^2 - 8x + 8y \;= \;0$

Here is what I got so far: . $A = 1,\;B = 2,\;C =1$

$\cot 2\theta \:= \:\frac{A - C}{B} \:=\:\frac{0}{2} \:=\:0$

$\cot2\theta \:=\:0\quad\Rightarrow\quad 2\theta \:=\:90^o \quad\Rightarrow\quad \theta \:=\:45^o$

$x \:= \:x'\cos45^o - y'\sin45^o \:=\:\frac{1}{\sqrt{2}}(x - y)$

$y \:= \:x'\sin45^o + y'\cos45^o \;=\;\frac{1}{\sqrt{2}}(x + y)$ . . . . all correct so far!

i got to this far and plug in to the original equation and simplied it.
But I don't know what to do after that.
Do you recognize the equation?

Substitute:
. $\left[\frac{1}{\sqrt{2}}(x-y)\right]^2 + 2\left[\frac{1}{\sqrt{2}}(x-y)\right]\left[\frac{1}{\sqrt{2}}(x+y)\right] + \left[\frac{1}{\sqrt{2}}(x + y)\right]^2$ $-8\left[\frac{1}{\sqrt{2}}(x-y)\right] + 8\left[\frac{1}{\sqrt{2}}(x + y)\right] \;=\;0$

. . . . $\frac{1}{2}x^2 - xy + \frac{1}{2}y^2 + x^2-y^2 + \frac{1}{2}x^2 + xy + \frac{1}{2}y^2
-4\sqrt{2} + 4\sqrt{2}y + 4\sqrt{2}x + 4\sqrt{2}y\;=\;0$

. . . . $x^2 + 4\sqrt{2}y \;=\;0\quad\Rightarrow\quad x^2 \:=\:-4\sqrt{2}y$

This is a parabola, opening upward, vertex at the origin.

3. Thank you so much Soroban! I got confused at the end when I tried to substitute it in the original problem, so I stopped and asked for help. .. I should have think more about this before asking for help. Again, thank you so much sir! Happy Thanksgiving!