Subsection 9.2

Second-order curves in a general Cartesian coordinate system (9.11-9.22)

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Problem 9.11

Prove that:

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(1)

the curve given by the equation x2+y2=1x^{2} + y^{2} = 1 is an ellipse

(2)

the curve given by the equation x2−y2=1x^{2} - y^{2} = 1 is a hyperbola

(3)

the curve given by the equation y2=xy^{2} = x is a parabola

(4)

the equation x2−y2=0x^{2} - y^{2} = 0 defines a pair of intersecting lines

(5)

the equation x2+y2=0x^{2} + y^{2} = 0 defines a single point

(6)

the equation y2−1=0y^{2} - 1 = 0 defines a pair of parallel lines

(7)

the equation y2=0y^{2} = 0 defines a single line (a pair of coincident lines)

(8)

the curve given by the equation xy=1xy = 1 is a hyperbola

(9)

the curve given by the equation y=x+1xy = x + \frac{1}{x} is a hyperbola.

Problem 9.12

Prove that the curve given, in a general Cartesian coordinate system, by the equation Ax2+2Bxy+Cy2+2Dx+2Ey+F=0Ax^{2} + 2Bxy + Cy^{2} + 2Dx + 2Ey + F = 0, is a hyperbola if and only if δ<0\delta < 0 and Δ≠0\Delta \neq 0, an ellipse if and only if δ>0\delta > 0 and SΔ<0S\Delta < 0, a parabola if and only if δ=0\delta = 0 and Δ≠0\Delta \neq 0.

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Problem 9.13

Determine the type of the curve of the second order given by the equation:

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(1)

(3x−4y)2−5(x+2y−1)2=1(3x - 4y)^{2} - 5(x + 2y - 1)^{2} = 1

(2)

(12x−17y−6)2+(17y+5x+1)2=1(12x - 17y - 6)^{2} + (17y + 5x + 1)^{2} = 1

(3)

(x−y−3)(x+y+3)=4(x - y - 3)(x + y + 3) = 4

(4)

(4x+3y−1)2+(4x+3y+2)2=5(4x + 3y - 1)^{2} + (4x + 3y + 2)^{2} = 5

(5)

17x2−2xy+y2−3x−y−3=017x^{2} - 2xy + y^{2} - 3x - y - 3 = 0

(6)

4x2+28xy+49y2−3x−15y+2=04x^{2} + 28xy + 49y^{2} - 3x - 15y + 2 = 0

(7)

4x2−12xy+8y2−15x+25y+14=04x^{2} - 12xy + 8y^{2} - 15x + 25y + 14 = 0

(8)

2x2+2xy+5y2−2y+4=02x^{2} + 2xy + 5y^{2} - 2y + 4 = 0

(9)

2x2−5xy−3y2+9x+y+4=02x^{2} - 5xy - 3y^{2} + 9x + y + 4 = 0

(10)

x2+10xy+25y2+2x+10y−3=0x^{2} + 10xy + 25y^{2} + 2x + 10y - 3 = 0

(11)

5x2−16xy+13y2+6x−10y+2=05x^{2} - 16xy + 13y^{2} + 6x - 10y + 2 = 0

(12)

x2−4xy+4y2+4x−8y+5=0x^{2} - 4xy + 4y^{2} + 4x - 8y + 5 = 0

(13)

x2−8xy+16y2+6x−24y+9=0x^{2} - 8xy + 16y^{2} + 6x - 24y + 9 = 0.

Problem 9.14

Find the equation and determine the type of the curve of the second order passing through 5 points given by their coordinates:

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(1)

(−1,−1),(1,0),(0,1),(3,2),(2,3)(- 1, - 1), (1, 0), (0, 1), (3, 2), (2, 3)

(2)

(1,1),(1,0),(0,1),(3,2),(2,3)(1, 1), (1, 0), (0, 1), (3, 2), (2, 3)

(3)

(−1,0),(1,0),(0,1),(3,2),(2,3)(- 1, 0), (1, 0), (0, 1), (3, 2), (2, 3)

(4)

(−3,0),(1,0),(0,1),(3,2),(2,3)(- 3, 0), (1, 0), (0, 1), (3, 2), (2, 3)

(5)

(−1,1),(0,1),(2,3),(−2,−1),(3,4)(- 1, 1), (0, 1), (2, 3), (- 2, - 1), (3, 4)

(6)

(1,0),(0,1),(1/4,1/4),(4/9,1/9),(1/9,4/9)(1, 0), (0, 1), (1 / 4, 1 / 4), (4 / 9, 1 / 9), (1 / 9, 4 / 9).

Problem 9.15

Investigate the dependence of the type of the curve of the second order on the parameter:

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(1)

4x2+2λxy+y2=14x^{2} + 2\lambda xy + y^{2} = 1

(2)

λ(x2+y2)−10xy+x+y+4=0\lambda \left(x^{2} + y^{2}\right) - 10xy + x + y + 4 = 0

(3)

x2−2xy+y2(λ−1)+2λ(x−y+1)=0x^{2} - 2xy + y^{2}(\lambda - 1) + 2\lambda (x - y + 1) = 0

(4)

λx2−2xy+2y2−2x+2y−1=0\lambda x^{2} - 2xy + 2y^{2} - 2x + 2y - 1 = 0.

Problem 9.16

What types of curves of the second order can be given by the equation:

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(1)

(A1x+B1y+C1)2=A2x+B2y+C2\left(A_{1} x + B_{1} y + C_{1}\right)^{2} = A_{2} x + B_{2} y + C_{2}

(2)

(A1x+B1y+C1)2+(A2x+B2y+C2)2=1\left(A_{1} x + B_{1} y + C_{1}\right)^{2} + \left(A_{2} x + B_{2} y + C_{2}\right)^{2} = 1

(3)

(A1x+B1y+C1)2−(A2x+B2y+C2)2=1\left(A_{1} x + B_{1} y + C_{1}\right)^{2} - \left(A_{2} x + B_{2} y + C_{2}\right)^{2} = 1

(4)

(A1x+B1y+C1)(A2x+B2y+C2)=1\left(A_{1} x + B_{1} y + C_{1}\right)\left(A_{2} x + B_{2} y + C_{2}\right) = 1

(5)

(A1x+B1y+C1)(A2x+B2y+C2)=0\left(A_{1} x + B_{1} y + C_{1}\right)\left(A_{2} x + B_{2} y + C_{2}\right) = 0?

Problem 9.17

Find the equations of the asymptotes of the hyperbola (It is assumed that A1B2−A2B1≠0A_{1} B_{2} - A_{2} B_{1} \neq 0).

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(1)

(A1x+B1y+C1)2−(A2x+B2y+C2)2=1\left(A_{1} x + B_{1} y + C_{1}\right)^{2} - \left(A_{2} x + B_{2} y + C_{2}\right)^{2} = 1

(2)

(A1x+B1y+C1)(A2x+B2y+C2)=1\left(A_{1} x + B_{1} y + C_{1}\right)\left(A_{2} x + B_{2} y + C_{2}\right) = 1.

Problem 9.18

Without using equations (6) from the introduction to the present chapter, prove that the origin is a center of symmetry of a curve of the second order if and only if the equation of the curve contains no terms with first powers of the variables xx and yy. Based on this statement, derive equations (6) for the coordinates of the center of a curve of the second order.

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Problem 9.19

Verify that the given curve of the second order is central. Find the coordinates of the center and eliminate, in the equation, the terms of the first degree by shifting the origin to the center

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(1)

x2−8xy+17y2+8x−38y+24=0x^{2} - 8xy + 17y^{2} + 8x - 38y + 24 = 0

(2)

5x2+xy−4x−y−1=05x^{2} + xy - 4x - y - 1 = 0

(3)

8x2−24xy+16y2+3x−7y−2=08x^{2} - 24xy + 16y^{2} + 3x - 7y - 2 = 0.

Problem 9.20

Prove that the curve Ax2+2Bxy+Cy2+2Dx+2Ey+F=0Ax^{2} + 2Bxy + Cy^{2} + 2Dx + 2Ey + F = 0 has a unique center of symmetry if and only if δ≠0\delta \neq 0.

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Problem 9.21

Prove that the set of centers of symmetry of a curve of the second order is either empty, or consists of a single point, or is a straight line.

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Problem 9.22
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(1)

Prove that the set of centers of symmetry of an algebraic curve is either empty, or consists of a single point, or is a straight line.

(2)

Prove that the set of centers of symmetry of an arbitrary set of points in the plane is either empty, or consists of a single point, or is infinite.

(3)

Give an example of a continuous curve whose set of centers of symmetry is infinite, but is not a straight line.