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