Answer:
The answer is ellipse; 3x² + y² + 6x - 6y + 3 = 0
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:
* 3x² + y² = 9
∵ A = 3 , B = 0 , C = 1
∴ B² - 4 AC = (0)² - 4(3)(1) = -12
∴ B² - 4AC < 0
∴ The graph is ellipse or circle
* If A and C are nonzero, have the same sign, and are not
equal to each other, then the graph is an ellipse.
* If A and C are equal and nonzero and have the same
sign, then the graph is a circle.
∵ A and C have same signs with different values
∴ It is an ellipse
* Now lets study T(-1 , 3), that means the graph will translate
1 unit to the left and 3 units up
∴ x will be (x - -1) = (x + 1) and y will be (y - 3)
* Lets substitute the x by ( x + 1) and y by (y - 3) in the equation
∴ 3(x + 1)² + (y - 3)² = 9
* Use the foil method
∴ 3(x² + 2x + 1) + (y² - 6y + 9) = 9
* Open the brackets
∴ 3x² + 6x + 3 + y² - 6y + 9 = 9
* Collect the like terms
∴ 3x² + y² + 6x - 6y + 12 = 9
∴ 3x² + y² + 6x - 6y + 12 - 9 = 0
∴ 3x² + y² + 6x - 6y + 3 = 0
* The answer is ellipse of equation 3x² + y² + 6x - 6y + 3 = 0