Pre calc
posted by Rebekah .
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
thanks

the axes are rotated through θ where
tan2θ = B/(AC) = 24/(3441) = 24/7
cos2θ = 7/25
cosθ = √((1+7/25)/2) = √(32/50) = 4/5
sinθ = √((17/25)/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/(AC) = 4/3
cos2θ = 3/5
cosθ = √((13/5)/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.