Question

Identify the graph of 2x^2+2y^=9 for theta=30º and write and equation of the translated or rotated graph in general form.

Answer 1

Answer:

The answer is D

Good luck on the Ed-genuity test

Answer 2

Answer:

The answer is circle; (x')² + (y')² - 4 = 0

Step-by-step explanation:

* At first lets talk about the general form of the conic equation

- Ax² + Bxy + Cy²  + Dx + Ey + F = 0

∵ B² - 4AC < 0 , if a conic exists, it will be either a circle or an ellipse.

∵ B² - 4AC = 0 , if a conic exists, it will be a parabola.

∵ B² - 4AC > 0 , if a conic exists, it will be a hyperbola.

* Now we will study our equation:

* 2x² + 2y² = 8

∵ A = 2 , B = 0 , C = 2

∴ B² - 4AC = (0) - 4(2)(2) = -16 < 0

∵ B² - 4AC < 0

∴  it will be either a circle or an ellipse

* Lets use this note to chose the correct figure

- If A and C are equal and nonzero and have the same sign,

 then the graph is a circle.

- If A and C are nonzero, have the same sign, and are not equal

 to each other, then the graph is an ellipse.

∵ A = 2 and C = 2

∴ The graph is a circle.

∵ D and E = 0

∴ The center of the circle is the origin (0 , 0)

∵ Ф = 30°

∴ The point (x , y) will be (x' , y')

- Where x = x'cosФ - y' sinФ and y = x'sinФ + y'cosФ

∴ x = x'cos(30°) - y'sin(30°)

∴ y = x'sin(30°) + y'cos(30°)

∴ x = (√3/2)x' - (1/2)y' and y = (1/2)x' + (√3/2)y'

∴ $x=\frac{\sqrt{3}x'-y'}{2}$

∴ $y=\frac{x'+\sqrt{3}y'}{2}$

* Lets substitute x and y in the first equation

∴ $2(\frac{\sqrt{3}x'-y'}{2})^{2}+2(\frac{x'+\sqrt{3}y'}{2})^{2}=8$

* Use the foil method

∴ $2(\frac{3x'^{2}-2\sqrt{3}x'y'+y'^{2}}{4})+2(\frac{x'^{2}+2\sqrt{3}x'y'+3y'^{2}}{4})=8$

* Open the brackets

∴ $\frac{3x'^{2}-2\sqrt{3}x'y'+y'^{2}+x'^{2}+2\sqrt{3}x'y'+3y'^{2}}{2}=8$

* Collect the like terms

∴ $\frac{4x'^{2}+4y'^{2}}{2}=8$

* Simplify the fraction

∴ 2(x')² + 2(y')²= 8

* Divide each side by 2

∴ (x')² + (y')² = 4

∴ The equation of the circle is (x')² + (y')² = 4

* The general equation of the circle is (x')² + (y')² - 4 = 0  

 after rotation 30° about the origin

* Look to the graph

- The blue circle for the equation 2x² + 2y² = 8

- The blue circle for equation (x')² + (y')² - 4 = 0

* That is because the two circles have same centers and radii

- The green line is x' and the purple line is y'