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

B

Explanation:

edge

Answer 2

The answer is circle; 2(x')² + 2(y')² - 5x' - (5√3)y' - 6 = 0 ⇒ answer (b)

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=(√(3))/(2)x'+(1)/(2)y'=((√(3)x'+y')/(2))`

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

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

∵
`((x'-√(3)y')/(2))^(2)-5((x'-√(3)y')/(2))+((√(3)x'-y')/(2))^(2)=3`

* Lets use the foil method

∴
`((x'^(2) -2√(3)x'y'+3y'))/(4)-((5x'-5√(3)y'))/(2)+((√(3)x'+2√(3)x'y'+y'^(2)))/(4)`=3

* Make L.C.M

∴
`((x'^(2)-2√(3)x'y'+3y'^(2)))/(4)-((10x'-10√(3)y'))/(4)+((3x'^(2)+2√(3)x'y'+y'^(2)))/(4) =3`

* Open the brackets ∴
`(x'^(2)-2√(3)x'y'+3y'^(2)-10x'+10√(3)y'+3x'^(2)+2√(3)x'y'+y'^(2))/(4)=3`

* Collect the like terms

∴
`(4x'^(2)+4y'^(2)-10x'+10√(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