Find equation of the rotated graph: 7(x)^2 - 6(3)^1/2 xy + 13 (y)^2 = 16   rotated by 30 degrees.

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beckden | High School Teacher | (Level 1) Educator

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We are going to transform using the following equations, by theta=30^o

 

`x = x'costheta - y'sintheta = sqrt(3)/2 x' - 1/2 y' `

`y = x'sintheta + y'costheta = 1/2 x' + sqrt(3)/2 y'`

Substituting we get

`7(sqrt(3)/2 x' - 1/2 y')^2 - 6sqrt(3) (sqrt(3)/2 x' - 1/2y') (1/2 x' + sqrt(3)/2 y') + 13(1/2 x' + sqrt(3)/2 y')^2 = 16`

`7(sqrt(3)/2x' - 1/2 y')^2 = 7(3/4 x'^2 - (sqrt(3))/2x'y' + 1/4y'^2)`

`7(sqrt(3)/2x' - 1/2 y')^2 = 21/4 x^2 - (7sqrt(3))/2x'y' + 7/4y'^2`

 

`6sqrt(3)(sqrt(3)/2x'-1/2y')(1/2x'+sqrt(3)/2y') = 6sqrt(3)(sqrt(3)/4x'^2 + (3/4 - 1/4)x'y' - sqrt(3)/4 y'^2) `

 

`6sqrt(3)(sqrt(3)/2x'-1/2y')(1/2x'+sqrt(3)/2y') = 6sqrt(3)(sqrt(3)/4x'^2 + 1/2x'y' - sqrt(3)/4 y'^2) `

 

`6sqrt(3)(sqrt(3)/2x'-1/2y')(1/2x'+sqrt(3)/2y') = 18/4x'^2 + (6sqrt(3))/2x'y' - (18/4)y'^2 `

 

`6sqrt(3)(sqrt(3)/2x'-1/2y')(1/2x'+sqrt(3)/2y') = 9/2x'^2 + 3sqrt(3)x'y' - 9/2y'^2 `

 

`13(1/2x' + sqrt(3)/2y')^2 = 13(1/4x'^2 + sqrt(3)/2x'y' + 3/4y'^2) `

 

`13(1/2x' + sqrt(3)/2y')^2 = 13/4x'^2 + (13sqrt93)/2x'y' + 39/4y'^2 `

 

`21/4 x^2 - (7sqrt(3))/2x'y' + 7/4y'^2 - (9/2x'^2 + 3sqrt(3)x'y' - 9/2y'^2) + 13/4x'^2 + (13sqrt93)/2x'y' + 39/4y'^2 = 16 `

Changing the subtraction to an addition my changing the sign of what I subtract we get

 

`21/4 x^2 - (7sqrt(3))/2x'y' + 7/4y'^2 - 9/2x'^2 - 3sqrt(3)x'y' + 9/2y'^2 + 13/4x'^2 + (13sqrt(3))/2x'y' + 39/4y'^2 = 16 `

Combining like terms

 

`(21/4 - 9/2 + 13/4)x'^2 + (-(7sqrt(3)/2 - 3sqrt(3) + (13sqrt(3))/2)) x'y' +(7/4 + 9/2 + 39/4)y'^2 = 16 `

 

`16/4x^2 + (-7/2 - 3 + 13/2)sqrt(3)x'y' + (64/4)y'^2 = 16 `

`4x'^2 - 0x'y' + 16y'^2 = 16`

So we get the rotated equation as

`4x^2 + 16y^2 = 16` or in standard form `x^2/4 + y^2/1 = 1`

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