MATH SOLVE

3 months ago

Q:
# Identify the graph of x^2-5x+y^2=3 for theta π/3 and write and equation of the translated or rotated graph in general form.

Accepted Solution

A:

Answer:The answer is circle; 2(x')² + 2(y')² - 5x' - (5√3)y' - 6 = 0 ⇒ answer (b)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:* x² - 5x + y² = 3∵ A = 1 , B = 0 , C = 1∴ B² - 4AC = (0) - 4(1)(1) = -4 < 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 = 1 an d C = 1∴ The graph is a circle.* To find its center of the circle lets use∵ h = -D/2A and k = -E/2A∵ A = 1 and D = -5 , E = 0∴ h = -(-5)/2(1) = 2.5 and k = 0∴ The center of the circle is (2.5 , 0)* Now lets talk about the equation of the circle and angle Ф ∵ Ф = π/3- That means the graph of the circle will transformed by angle = π/3- The point (x , y) will be (x' , y'), where* x = x'cos(π/3) - y'sin(π/3) , y = x'sin(π/3) + y'cos(π/3)∵ cos(π/3) = 1/2 and sin(π/3) = √3/2∴ [tex]y=\frac{\sqrt{3}}{2}x'+\frac{1}{2}y'=(\frac{\sqrt{3}x'+y'}{2})[/tex]∴ [tex]x=\frac{1}{2}x'-\frac{\sqrt{3}}{2}y'=(\frac{x'-\sqrt{3}y'}{2})[/tex]* Lets substitute x and y in the equation x² - 5x + y² = 3∵ [tex](\frac{x'-\sqrt{3}y'}{2})^{2}-5(\frac{x'-\sqrt{3}y'}{2})+(\frac{\sqrt{3}x'-y'}{2})^{2}=3[/tex]* Lets use the foil method∴ [tex]\frac{(x'^{2} -2\sqrt{3}x'y'+3y')}{4}-\frac{(5x'-5\sqrt{3}y')}{2}+\frac{(\sqrt{3}x'+2\sqrt{3}x'y'+y'^{2})}{4}[/tex]=3* Make L.C.M∴ [tex]\frac{(x'^{2}-2\sqrt{3}x'y'+3y'^{2})}{4}-\frac{(10x'-10\sqrt{3}y')}{4}+\frac{(3x'^{2}+2\sqrt{3}x'y'+y'^{2})}{4} =3[/tex]* Open the brackets ∴[tex]\frac{x'^{2}-2\sqrt{3}x'y'+3y'^{2}-10x'+10\sqrt{3}y'+3x'^{2}+2\sqrt{3}x'y'+y'^{2}}{4}=3[/tex]* Collect the like terms∴ [tex]\frac{4x'^{2}+4y'^{2}-10x'+10\sqrt{3}y'}{4}=3[/tex]* Multiply both sides by 4∴ 4(x')² + 4(y')² - 10x' + (10√3)y' = 12* Divide both sides by 2∴ 2(x')² + 2(y')² - 5x' + (5√3)y' = 6∵ h = -D/2A and k = E/2A∵ A = 2 and D = -5 , E = 5√3∴ h = -(-5)/2(2) = 5/4 =1.25∵ k = (5√3)/2(2) = (5√3)/4 = 1.25√3∴ The center of the circle is (1.25 , 1.25√3)∵ The center of the first circle is (2.5 , 0)∵ The center of the second circle is (1.25 , 1.25√3)∴ The circle translated Left and up * 2(x')² + 2(y')² - 5x' - (5√3)y' - 6 = 0∴ The answer is circle; 2(x')² + 2(y')² - 5x' - (5√3)y' - 6 = 0* Look to the graph- the purple circle for the equation x² - 5x + y² = 3- the black circle for the equation (x')² + (y')² - 5x' - 5√3y' - 6 = 0