Posted by Rebekah on Sunday, May 5, 2013 at 2:18pm.
Write the equation in standard ellipse form and graph.
34 x^2 -24xy +41y^2 +41 y^2 -25= 0.
i got x= (4 x prime- 3 y prime)/5 and y=(3x prime +4y prime)/5
but cant get the correct math when I plug that into the original equation.
Also I need help finding the equation for
x^2 +4xy+4y^2 +5root5y+5=0 because i think u are suppose to find the square because of the 5root5y in there but am not sure
- Pre calc - Steve, Sunday, May 5, 2013 at 6:21pm
the axes are rotated through θ where
tan2θ = B/(A-C) = -24/(34-41) = 24/7
cos2θ = 7/25
cosθ = √((1+7/25)/2) = √(32/50) = 4/5
sinθ = √((1-7/25)/2) = √(18/50) = 3/5
So, you are correct that
x = (4x' - 3y')/5
y = (3x' + 4y')/5
x^2 = (16x'^2 - 24x'y' + 9y'^2)/25
xy = (12x'^2 + 7x'y' - 12 y'^2)/25
y^2 = (9x'^2 + 24x'y' + 16y'^2)/25
Now it appears there's a typo. Why is 41y^2 there twice?
For the 2nd I think the √5 is there on purpose. It just means that the coefficients aren't all integers.
tan2θ = B/(A-C) = -4/3
cos2θ = -3/5
cosθ = √((1-3/5)/2) = -1/√5
sinθ = √((1+3/5)/2) = -2/√5
x = 1/√5 (-x' + 2y')
y = 1/√5 (-2x' - y')
x^2 = 1/5 (x'^2 - 4x'y' + 4y'^2)
xy = 1/5 (-2x'^2 - 3x'y' - 2y'^2)
y^2 = 1/5 (4x'^2 + 4x'y' + y'^2)
x^2+4xy+4y^2 + 5√5 y + 5
= 9x'^2 + y'^2 - 10x' - 5y' + 5
= 9(x'^2 - 10/9 x' + 25/81) + (y'^2 - 5y' + 25/4) + 5 - 25/9 - 25/4
9(x' - 5/9)^2 + (y' - 5/2)^2 = 145/36
Hmmm. Better check my algebra. Wolframalpha says it's a parabola, not an ellipse.
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