Question

Rewrite the equation in a rotated x'y'-system without an x'y' term. Express the equation involving x' and y' in the standard form of a conic section.

31x2 + 10xy + 21y2 -144 = 0

Answer 1

angle = 30 so x = (sqrt3x - y)/2 y = (x + sqrt3y)/2

Answer 2

so i plg in x and y and i now have this long really confusing equation that im having trouble simplifying.....
did i even take the right steps?

Answer 3

Let me see if I can remember how to do...

We have Ax^2 + Bxy + Cy^2 +F = 0 whre

A = 31 B = 10 C = 21 D = -144

A' = A cos^2 theta + B sin theta cos theta + C sin^s theta

C' = A sin^2 theta - B sin theta cos theta + C cos^2 theta

F' = F and where Tan 2 theta = (B/(A-C))

Answer 4

B = 10sqrt 3
cot 2 theta = ((A - C)/B)
cot 2 theta = 1/sqrt 3
= 60 degrees/ 2 = 30 degrees

Answer 5

atleast that's the way ive learned it so far.
and then x = x'cos theta - y'sin theta
y = x'sin theta + y'cos theta

Answer 6

cot 2 theta = ((A - C)/B)
cot 2 theta = 1/sqrt 3
= 60 degrees/ 2 = 30 degrees

This is the same thing as mine.......

Answer 7

oh ok. but then what's next do i plug in those x and y values?

Answer 8

You need to do this (in my way)

A' = A cos^2 theta + B sin theta cos theta + C sin^s theta

C' = A sin^2 theta - B sin theta cos theta + C cos^2 theta

F' = F

and you can get all the trigonometry from a triangle given by Tan 2 theta = (B/(A-C)) = 1

Where do you get 30 degrees?

Answer 9

A-C/ B is 1 right?

Answer 10

i got 1/ sqrt3

Answer 11

31-21/10 ?

Answer 12

i included the sqrt3 so that B = 10(sqrt3)

Answer 13

What sqrt 3?

Answer 14

omg! i must have written that down on my paper different.
you're right there's no sqrt3

that changes everything! haha

Answer 15

OK, let me know if it still doesn't work out...:-)

Answer 16

ok thank you