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'
∴ [tex]x=\frac{\sqrt{3}x'-y'}{2}[/tex]
∴ [tex]y=\frac{x'+\sqrt{3}y'}{2}[/tex]
* Lets substitute x and y in the first equation
∴ [tex]2(\frac{\sqrt{3}x'-y'}{2})^{2}+2(\frac{x'+\sqrt{3}y'}{2})^{2}=8[/tex]
* Use the foil method
∴ [tex]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[/tex]
* Open the brackets
∴ [tex]\frac{3x'^{2}-2\sqrt{3}x'y'+y'^{2}+x'^{2}+2\sqrt{3}x'y'+3y'^{2}}{2}=8[/tex]
* Collect the like terms
∴ [tex]\frac{4x'^{2}+4y'^{2}}{2}=8[/tex]
* 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'
