Question

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.

Answer 1

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

∴ $y=\frac{\sqrt{3}}{2}x'+\frac{1}{2}y'=(\frac{\sqrt{3}x'+y'}{2})$

∴ $x=\frac{1}{2}x'-\frac{\sqrt{3}}{2}y'=(\frac{x'-\sqrt{3}y'}{2})$

* Lets substitute x and y in the equation x² - 5x + y² = 3

∵ $(\frac{x'-\sqrt{3}y'}{2})^{2}-5(\frac{x'-\sqrt{3}y'}{2})+(\frac{\sqrt{3}x'-y'}{2})^{2}=3$

* Lets use the foil method

∴ $\frac{(x'^{2} -2\sqrt{3}x'y'+3y')}{4}-\frac{(5x'-5\sqrt{3}y')}{2}+\frac{(\sqrt{3}x'+2\sqrt{3}x'y'+y'^{2})}{4}$=3

* Make L.C.M

∴ $\frac{(x'^{2}-2\sqrt{3}x'y'+3y'^{2})}{4}-\frac{(10x'-10\sqrt{3}y')}{4}+\frac{(3x'^{2}+2\sqrt{3}x'y'+y'^{2})}{4} =3$

* Open the brackets ∴$\frac{x'^{2}-2\sqrt{3}x'y'+3y'^{2}-10x'+10\sqrt{3}y'+3x'^{2}+2\sqrt{3}x'y'+y'^{2}}{4}=3$

* Collect the like terms

∴ $\frac{4x'^{2}+4y'^{2}-10x'+10\sqrt{3}y'}{4}=3$

* 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