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Lacey asked in Science & MathematicsMathematics · 7 years ago

# Help with rotation of axes?

~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.

24xy - 7y^2 + 36 = 0

My teacher did a poor job of explaining this section and I have no clue what to do! Any help will be greatly appreciated! :)

Relevance
• 7 years ago

If A is the rotation angle of the new co-ordinate system, we can express the current x, y co-ordinates in terms of the new co-ordinate system as:

x = x' cos A - y' sin A

y = x' sin A + y' cos A

Substituting these values into 24xy - 7y^2 + 36 = 0, we get:

24 (x' cos A - y' sin A)(x' sin A + y' cos A)

- 7 (x' sin A + y' cos A)^2 + 36

= 24 [x'^2 (sin A) (cos A) + (x'y')(cos^2 A - sin^2 A) - (y'^2) (sin A) (cos A)]

- 7 [x'^2 sin^2 A + (2x'y')(sin A) (cos A) + (y'^2 cos^2 A)

+ 36

Collecting x'^2, x'y' and y'^2 terms:

= x'^2 [24 (sin A) (cos A) - 7 sin^2 A]

+ x'y'[(cos^2 A - sin^2 A) - 14(sin A) (cos A)]

+ y'^2 [-24 (sin A) (cos A) - 7 cos^2 A]

+ 36

But 2 (sin A) (cos A) = sin 2A,

so 24 (sin A) (cos A) = 12 sin 2A

Also cos 2A

= cos^2 A - sin^2 A

= 1 - sin^2 A - sin^2 A = 1 - 2 sin^2 A

so 1 - cos 2A = 2 sin^2 A

So - 7 sin^2 A = (-7/2)(1 - cos 2A)

so the coefficient of x'^2 is 12 sin 2A -(7/2)(1 - cos 2A)

Since cos 2A = cos^2 A - sin^2 A

= cos^2 A - (1 - cos^2 A)

= 2 cos^2 A - 1,

then 1 + cos 2A = 2 cos^2 A,

so -(7/2)(1 + cos 2A) = - 7 cos^2 A,

Then coefficient of y'^2 is:

-24 (sin A) (cos A) - 7 cos^2 A

= -12 sin 2A - (7/2)(1 + cos 2A)

The coefficient of x'y' is therefore

cos 2A - 7 sin 2A.

To easily identify the conic section, we want this coefficient to vanish, so we choose A accordingly. If we then divide by cos 2A, we get:

1 - 7 tan 2A = 0, so tan 2A = 1,

i.e. tan 2A = 1/7.

The equation in the new co-ordinate system becomes:

[12 sin 2A -(7/2)(1 - cos 2A)] x'^2

+ [0] x'y'

+ [-12 sin 2A - (7/2)(1 + cos 2A)] y'^2

+ 36 = 0

Dividing through by cos 2A:

[12 tan 2A -(7/2)(sec 2A - 1)]x'^2

+ [-12 tan 2A - (7/2)(sec 2A + 1)] y'^2

+ 36 sec 2A = 0

But tan 2A = 1/7,

and sec 2A = sqrt (sin^2 2A + cos^ 2A) / (cos 2A)

= sqrt (tan^2 2A + 1)

= sqrt (1/49 + 1) = sqrt (50/49) = sqrt(50)/7

So:

[(12/7) +(7/2)(1 - sqrt(50)/7)]x'^2

- [(12/7) + (7/2)(1 + sqrt(50)/7] y'^2

+ 36 sqrt(50)/7 = 0

Multiplying by 14:

[24 +7(7 - sqrt(50))]x'^2

- [24 + 7(7 + sqrt(50)] y'^2

+ 72 sqrt(50) = 0

=[24+49-7 sqrt(50)]x'^2

- [24 + 49+7 sqrt(50)] y'^2

+ 72 sqrt(50)

=[73-7 sqrt(50)]x'^2

- [73+7 sqrt(50)] y'^2

+ 72 sqrt(50)

Define u=73-7 sqrt(50), v=73+7 sqrt(50), w=72 sqrt(50).

Then this can be expressed as:

u x'^2 - v y'^2 + w = 0.

Dividing through by w and rearranging:

1 = (v/w) y'^2 - (u/w) x'^2.

This is consistent with the form:

y'^2/a^2 - x'^2/b^2 = 1

which is a hyperbola.