Question

Help with Rotation of Axes

Answer 1

Determine the angle of rotation of the conic section given by: x2 + xy + y2 = 10 (round your answer to the nearest tenth of a degree).

Answer 2

cot(2Theta)=(1-1)/1 ?

Answer 3

the conic section is x^2+xy+ y^2=10.. correct?

Answer 4

Yes

Answer 5

okay so you are probably familiar with the basic Ax^2 + Cy^2 + Dx + Ey + F = 0 conic section equation

Answer 6

Yes

Answer 7

ight so

general equation of ellipse:
Ax^2+Bxy+Cy^2+Dx+Ey+F=0

if:
4AC−B^2>0

The trick is to eliminate B so that the xy term goes away just like you were doing.

cot(2θ) = A-C/B

Answer 8

so then plugging in the equation you should get something like

θ=π/4±πn/2

Answer 9

Sine we were given the terms did I put them incorrectly? it would be 0/1 added together s:

Answer 10

no you did it right I believe. (1-1)/1!

Answer 11

2Theta= cot^-1(0)?

Answer 12

(pi/2) /2

Answer 13

Theta=pi/4 or 45 degrees

Answer 14

Correct you should get 2theta=arccos(0) which simplify's to θ=π/4

Answer 15

use the rotation transformations \(x=x'\cos\theta-y'\sin\theta\) and \(y=x'\sin\theta+y'\cos\theta\) so $$x^2=x'^2\cos^2\theta+y'^2\sin^2\theta-2x'y'\cos\theta\sin\theta\\y^2=x'^2\sin^2\theta+y'^2\cos^2\theta+2x'y'\cos\theta\sin\theta$$ and lastly $$xy=x'^2\cos\theta\sin\theta+x'y'\cos^2\theta-x'y'\sin^2\theta-y'^2\cos\theta\sin\theta$$ we find: $$x^2+y^2+xy=10\\(x'^2\cos^2\theta+y'^2\sin^2\theta-2x'y'\cos\theta\sin\theta)+(x'^2\sin^2\theta+y'^2\cos^2\theta+2x'y'\cos\theta\sin\theta)\\\quad+(x'^2\cos\theta\sin\theta+x'y'\cos^2\theta-x'y'\sin^2\theta-y'^2\cos\theta\sin\theta)=10\\(x'^2+y'^2)(\cos^2\theta+\sin^2\theta)+x'y'(\cos^2\theta-\sin^2\theta)+(x'^2-y'^2)\cos\theta\sin\theta=10$$for this to reduce with no mixed terms we must have \(\cos^2\theta-\sin^2\theta=0\), so $$\tan^2\theta=1\implies\theta=\frac{(2n+1)\pi}4$$for integer \(n\), so the conic is rotated by \(\pi/4\)