We are given several equations and asked to identify the conic or limiting form represented by each equation. We will use the method of completing the square to rewrite the equations into a standard form, which will allow us to identify the conic sections.

GeometryConic SectionsCirclesHyperbolasCompleting the Square
2025/5/30

1. Problem Description

We are given several equations and asked to identify the conic or limiting form represented by each equation. We will use the method of completing the square to rewrite the equations into a standard form, which will allow us to identify the conic sections.

2. Solution Steps

Problem 1: x2+y22x+2y+1=0x^2 + y^2 - 2x + 2y + 1 = 0
Complete the square for the x terms and y terms.
(x22x)+(y2+2y)+1=0(x^2 - 2x) + (y^2 + 2y) + 1 = 0
(x22x+1)+(y2+2y+1)+111=0(x^2 - 2x + 1) + (y^2 + 2y + 1) + 1 - 1 - 1 = 0
(x1)2+(y+1)21=0(x - 1)^2 + (y + 1)^2 - 1 = 0
(x1)2+(y+1)2=1(x - 1)^2 + (y + 1)^2 = 1
This is the equation of a circle with center (1,1)(1, -1) and radius 11.
Problem 2: x2+y2+6x2y+6=0x^2 + y^2 + 6x - 2y + 6 = 0
Complete the square for the x terms and y terms.
(x2+6x)+(y22y)+6=0(x^2 + 6x) + (y^2 - 2y) + 6 = 0
(x2+6x+9)+(y22y+1)+691=0(x^2 + 6x + 9) + (y^2 - 2y + 1) + 6 - 9 - 1 = 0
(x+3)2+(y1)24=0(x + 3)^2 + (y - 1)^2 - 4 = 0
(x+3)2+(y1)2=4(x + 3)^2 + (y - 1)^2 = 4
This is the equation of a circle with center (3,1)(-3, 1) and radius 22.
Problem 9: 3x2+3y26x+12y+60=03x^2 + 3y^2 - 6x + 12y + 60 = 0
Divide by 3: x2+y22x+4y+20=0x^2 + y^2 - 2x + 4y + 20 = 0
Complete the square for the x terms and y terms.
(x22x)+(y2+4y)+20=0(x^2 - 2x) + (y^2 + 4y) + 20 = 0
(x22x+1)+(y2+4y+4)+2014=0(x^2 - 2x + 1) + (y^2 + 4y + 4) + 20 - 1 - 4 = 0
(x1)2+(y+2)2+15=0(x - 1)^2 + (y + 2)^2 + 15 = 0
(x1)2+(y+2)2=15(x - 1)^2 + (y + 2)^2 = -15
Since the sum of squares cannot be negative, there are no real solutions. This represents no conic section.
Problem 10: 4x24y22x+2y+1=04x^2 - 4y^2 - 2x + 2y + 1 = 0
4x22x4y2+2y+1=04x^2 - 2x - 4y^2 + 2y + 1 = 0
4(x212x)4(y212y)+1=04(x^2 - \frac{1}{2}x) - 4(y^2 - \frac{1}{2}y) + 1 = 0
4(x212x+116)4(y212y+116)+14(116)+4(116)=04(x^2 - \frac{1}{2}x + \frac{1}{16}) - 4(y^2 - \frac{1}{2}y + \frac{1}{16}) + 1 - 4(\frac{1}{16}) + 4(\frac{1}{16}) = 0
4(x14)24(y14)2+1=04(x - \frac{1}{4})^2 - 4(y - \frac{1}{4})^2 + 1 = 0
4(x14)24(y14)2=14(x - \frac{1}{4})^2 - 4(y - \frac{1}{4})^2 = -1
(y14)2(x14)2=14(y - \frac{1}{4})^2 - (x - \frac{1}{4})^2 = \frac{1}{4}
This is a hyperbola.

3. Final Answer

1. Circle

2. Circle

3. No conic section

4. Hyperbola

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