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/(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.
11,13,17,23,29 Is Composite Nuber??
To start, let's rearrange and simplify the first equation:
34x^2 - 24xy + 41y^2 + 41y^2 - 25 = 0
Combine like terms:
34x^2 - 24xy + 82y^2 - 25 = 0
Now, let's focus on completing the square to convert the equation into standard ellipse form:
1. Start by grouping the x and y terms separately:
(34x^2 - 24xy) + (82y^2) = 25
2. Factor out the coefficient of x^2 from the x terms and the coefficient of y^2 from the y terms:
2(17x^2 - 12xy) + 82y^2 = 25
3. Now, we want to complete the square for both the x and y terms.
Completing the square for x:
- Take half of the coefficient of xy (in this case, -12) and square it: (-12/2)^2 = 36
- Add and subtract the result inside the parentheses:
2(17x^2 - 12xy + 36 - 36) + 82y^2 = 25
Simplify this part:
2((sqrt(17)x - 6y)^2 - 36) + 82y^2 = 25
Completing the square for y:
- Take half of the coefficient of y^2 (in this case, 82) and square it: (82/2)^2 = 3364
- Add and subtract the result inside the parentheses:
2((sqrt(17)x - 6y)^2 - 36) + 3364y^2 - 3364 = 25
Simplify this part:
2((sqrt(17)x - 6y)^2 + 3364y^2 - 3392) = 25
Divide by the constant term to isolate the term on the other side:
((sqrt(17)x - 6y)^2)/[(25/2)] + (3364y^2 - 3392)/25 = 1
Simplify the constants:
((sqrt(17)x - 6y)^2)/(25/2) + (3364/25) * (y^2 - 1) = 1
Now we have the equation in standard ellipse form:
((sqrt(17)x - 6y)^2)/(25/2) + (67.36/25) * (y^2 - 1) = 1
To graph this equation, you can plot the center, vertices, co-vertices, and axes using the standard ellipse form.
Regarding the second equation, x^2 + 4xy + 4y^2 + 5*sqrt(5)y + 5 = 0, it is not primarily an ellipse form. However, you correctly noted the term 5*sqrt(5)y, which suggests trying to complete the square.
The equation can be rewritten as:
(x^2 + 2xy + y^2) + (3xy + 3y^2 + 5*sqrt(5)y + 5) = 0
Now, let's complete the square part by part:
Completing the square for x and y terms:
(x + y)^2 + (3xy + 3y^2 + 5*sqrt(5)y + 5) = 0
Take half of the coefficient of xy (in this case, 3) and square it: (3/2)^2 = 9/4
Add and subtract inside the parentheses:
(x + y)^2 + (3xy + 9/4 + 3y^2 + 5*sqrt(5)y + 5 - 9/4) = 0
Simplify this part:
(x + y)^2 + (3xy + 3y^2 + 5*sqrt(5)y + 20/4) = 0
Now, rewrite the equation:
(x + y)^2 + 3(y + (sqrt(5)/2))^2 + (20/4) - (9/4) = 0
Simplify the constants:
(x + y)^2 + 3(y + (sqrt(5)/2))^2 + (11/2) = 0
This equation represents the equation of an ellipse. However, unlike the standard ellipse form, the coefficients of both the x and y squared terms are different.