Section 1.6

Center, Diameters. Simplification of Equations of Second-Order Curves

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Problem 271

Let (x,y)(x,y) be the midpoint of a chord of the curve f(x,y,1)=0f(x,y,1)=0, and let (x+ξt,y+ηt)(x+\xi t, y+\eta t) and (x−ξt,y−ηt)(x-\xi t, y-\eta t) be the endpoints of the chord. Prove that

(∂f∂x)1ξ+(∂f∂y)1η=0. \left(\frac{\partial f}{\partial x}\right)_{1}\xi + \left(\frac{\partial f}{\partial y}\right)_{1}\eta = 0.
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Remark.
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The polynomial Ax2+Bxy+Cy2+Dx+Ey+FAx^{2}+Bxy+Cy^{2}+Dx+Ey+F, homogenized by introducing powers of the variable zz, becomes the ternary quadratic form f=f(x,y,z)=Ax2+Bxy+Cy2+Dxz+Eyz+Fz2f = f(x,y,z) = Ax^{2}+Bxy+Cy^{2}+Dxz+Eyz+Fz^{2}. Its partial derivatives with respect to xx, yy and zz are given by: ∂f∂x=2Ax+By+Dz\frac{\partial f}{\partial x} = 2Ax+By+Dz; ∂f∂y=Bx+2Cy+Ez\frac{\partial f}{\partial y} = Bx+2Cy+Ez; ∂f∂z=Dx+Ey+2Fz\frac{\partial f}{\partial z} = Dx+Ey+2Fz. Their values at z=1z=1 are denoted by (∂f∂x)1\left(\frac{\partial f}{\partial x}\right)_{1}, (∂f∂y)1\left(\frac{\partial f}{\partial y}\right)_{1} and (∂f∂z)1\left(\frac{\partial f}{\partial z}\right)_{1}. For brevity one also writes f(x,y,1)=Ax2+Bxy+Cy2+Dx+Ey+Ff(x,y,1) = Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F, f(x,y,0)=Ax2+Bxy+Cy2f(x,y,0) = Ax^{2}+Bxy+Cy^{2}.

Problem 272

Prove that the midpoints of the chords of the curve f(x,y,1)=0f(x,y,1)=0 parallel to the line xξ=yη\frac{x}{\xi } = \frac{y}{\eta } lie on the line (∂f∂x)1ξ+(∂f∂y)1η=0\left(\frac{\partial f}{\partial x}\right)_{1}\xi + \left(\frac{\partial f}{\partial y}\right)_{1}\eta = 0, or, in full form:

(2Ax+By+D)ξ+(Bx+2Cy+E)η=0. (2Ax+By+D)\xi + (Bx+2Cy+E)\eta = 0.

(The diameter conjugate to the given direction.)

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Problem 273

Show that when the origin is translated to the point (ξ,η)(\xi ,\eta ), the equation f(x,y,1)=0f(x,y,1)=0 passes into the equation

f(x1,y1,0)+∂f∂ξx1+∂f∂ηy1+f(ξ,η,1)=0. f(x_{1},y_{1},0) + \frac{\partial f}{\partial \xi }x_{1} + \frac{\partial f}{\partial \eta }y_{1} + f(\xi ,\eta ,1) = 0.
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Problem 274

If 4AC−B2≠04AC-B^{2} \neq 0, all the diameters pass through the point (ξ,η)(\xi ,\eta ) where the lines (∂f∂x)1=0\left(\frac{\partial f}{\partial x}\right)_{1}=0, (∂f∂y)1=0\left(\frac{\partial f}{\partial y}\right)_{1}=0 intersect. Show that when the origin is translated to the point chosen in this way, the equation f(x,y,1)=0f(x,y,1)=0 passes into the following one:

Ax12+Bx1y1+Cy12+12(Dξ+Eη+2F)=0. Ax_{1}^{2}+Bx_{1}y_{1}+Cy_{1}^{2} + \frac{1}{2}(D\xi +E\eta +2F) = 0.

Here Dξ+Eη+2FD\xi +E\eta +2F is the value of (∂f∂z)1\left(\frac{\partial f}{\partial z}\right)_{1} at x=ξx=\xi, y=ηy=\eta.

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Problem 275

Translate the origin to the center of the curve x2+3xy−2y2−7x−2y+16=0x^{2}+3xy-2y^{2}-7x-2y+16=0 and set up the new equation of the curve.

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Problem 276

Simplify the equation of the curve

x2+3xy−2y2−7x−2y+16=0, x^{2}+3xy-2y^{2}-7x-2y+16=0,

by translating the origin to the center.

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Problem 277

Simplify the equation of the curve

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

by translating the origin to that center of it whose abscissa equals one.

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Problem 278

Show that the line 7x+y+6=07x+y+6=0 passes through the centers of the curves

3x2−7xy−6y2+3x−9y+5=0 3x^{2}-7xy-6y^{2}+3x-9y+5=0

and

3x2−5xy+6y2+11x−17y+13=0. 3x^{2}-5xy+6y^{2}+11x-17y+13=0.
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Problem 279

Find the axes of symmetry of the ellipse

x2+xy−y2−x+y−1=0. x^{2}+xy-y^{2}-x+y-1=0.
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Problem 280

Find the foci (with respect to the old coordinate axes) of the hyperbola x2−6xy+y2−2x−2y+5=0x^{2}-6xy+y^{2}-2x-2y+5=0.

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Problem 281

Determine the eccentricity of the curve

x2−4xy−2y2+10x+4y=0. x^{2}-4xy-2y^{2}+10x+4y=0.
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Problem 282

Simplify the equation of the curve

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

by translating the origin to the center, and then rotating the axes through the appropriate angle.

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Problem 283

The same, for the curve x2+6xy+y2+8x+24y+39=0x^{2}+6xy+y^{2}+8x+24y+39=0.

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Problem 284

Simplify the equation of the parabola

x2−2xy+y2−2x−2y+1=0, x^{2}-2xy+y^{2}-2x-2y+1=0,

by rotating the axes so that the new equation contains no x1y1x_{1}y_{1}, and then translating the origin so that only two terms remain in the final equation.

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Problem 285

Prove that when 4AC−B2=04AC-B^{2}=0, A>0A>0 (so that one may set A=a2A=a^{2}), the identity holds

Ax2+Bxy+Cy2+Dx+Ey+F=(ax+by+h)2+αx+βy+γ, Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F = (ax+by+h)^{2}+\alpha x+\beta y+\gamma ,

where α=D−2ah\alpha = D-2ah, β=E−2bh\beta = E-2bh, γ=F−h2\gamma = F-h^{2}.

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Problem 286

Prove that in the identity of the preceding problem, the number hh can be chosen so that the lines ax+by+h=0ax+by+h=0 and αx+βy+γ=0\alpha x+\beta y+\gamma =0 are mutually perpendicular, provided only that the numbers DD and EE are not proportional to the numbers aa and bb.

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Problem 287

Simplify the equation of the curve

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

using the identity:

x2+2xy+y2−9x+4=(x+y+h)2+αx+βy+γ, x^{2}+2xy+y^{2}-9x+4 = (x+y+h)^{2}+\alpha x+\beta y+\gamma ,

choosing hh so that the lines αx+βy+γ=0\alpha x+\beta y+\gamma =0 and x+y+h=0x+y+h=0 are mutually perpendicular. Here the formulas for the coordinate transformation have the form:

y1=x+y+h2,x1=αx+βy+γα2+β2. y_{1} = \frac{x+y+h}{\sqrt{2}}, \qquad x_{1} = \frac{\alpha x+\beta y+\gamma }{\sqrt{\alpha ^{2}+\beta ^{2}}}.
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Problem 288

By the same method, simplify the equation of the parabola

16x2+24xy+9y2−170x+310y−465=0. 16x^{2}+24xy+9y^{2}-170x+310y-465=0.
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Problem 289

Simplify the equation of the parabola

9x2+30xy+25y2+4x−16y+8=0. 9x^{2}+30xy+25y^{2}+4x-16y+8=0.
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Problem 290

For given AA, BB and CC, and a suitably chosen value of λ\lambda, the identity holds

Ax2+Bxy+Cy2−λ(x2+y2)=(αx+βy)2, Ax^{2}+Bxy+Cy^{2}-\lambda (x^{2}+y^{2}) = (\alpha x+\beta y)^{2},

where α\alpha and β\beta are real or imaginary coefficients. Prove that such values of λ\lambda are the roots of the equation

4λ2−4(A+C)λ+4AC−B2=0. 4\lambda ^{2}-4(A+C)\lambda +4AC-B^{2}=0.
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Problem 291

The equation

Ax2+Bxy+Cy2+Dx+Ey+F=0 Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F = 0

represents a pair of lines, real or imaginary, if and only if the discriminant Δ\Delta equals zero. Using this, prove that under any coordinate transformation the quantity Δ\Delta does not change.

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Remark.
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The discriminant Δ\Delta of the quadratic form in three variables f(x,y,z)=Ax2+Bxy+Cy2+Dxz+Eyz+Fz2f(x,y,z) = Ax^{2}+Bxy+Cy^{2}+Dxz+Eyz+Fz^{2} is defined as the determinant of the coefficients of xx, yy and zz in the formulas expressing its partial derivatives: Δ=∣2ABDB2CEDE2F∣\Delta = \begin{vmatrix} 2A & B & D \\ B & 2C & E \\ D & E & 2F \end{vmatrix}. Its minor δ=∣2ABB2C∣=4AC−B2\delta = \begin{vmatrix} 2A & B \\ B & 2C \end{vmatrix} = 4AC-B^{2} is the discriminant of the quadratic form in two variables Ax2+Bxy+Cy2Ax^{2}+Bxy+Cy^{2}.

Problem 292

Using the invariance of the quantities A+CA+C, 4AC−B24AC-B^{2} and the discriminant Δ\Delta, prove that when 4AC−B2≠04AC-B^{2} \neq 0, the equation

Ax2+Bxy+Cy2+Dx+Ey+F=0 Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0

reduces, after simplification, to the form

λ1x12+λ2x22=ΔB2−4AC. \lambda _{1}x_{1}^{2}+\lambda _{2}x_{2}^{2} = \frac{\Delta }{B^{2}-4AC}.

Here λ1\lambda_{1} and λ2\lambda_{2} are the roots of the equation λ2−2(A+C)λ+4AC−B2=0\lambda^{2}-2(A+C)\lambda +4AC-B^{2}=0.

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Problem 293

Prove that the equation Ax2+Bxy+Cy2+Dx+Ey+F=0Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0, when AC−D2=0AC-D^{2}=0 and Δ≠0\Delta \neq 0, can be reduced to the form

(A+C)x12=±−Δ2(A+C)y1. (A+C)x_{1}^{2} = \pm \sqrt{\frac{-\Delta }{2(A+C)}}y_{1}.
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Problem 294

Reduce the equation of the following line to its simplest form: 5x2+4xy+8y2−32x−56y+80=05x^{2}+4xy+8y^{2}-32x-56y+80=0.

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Problem 295

Reduce the equation of the following line to its simplest form: 5x2+8xy+5y2−18x−18y+9=05x^{2}+8xy+5y^{2}-18x-18y+9=0.

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Problem 296

Reduce the equation of the following line to its simplest form: 5x2+6xy+5y2−16x−16y+16=05x^{2}+6xy+5y^{2}-16x-16y+16=0.

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Problem 297

Reduce the equation of the following line to its simplest form: 5x2−4xy+5y2+4x−10y=3195x^{2}-4xy+5y^{2}+4x-10y=319.

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Problem 298

Reduce the equation of the following line to its simplest form: 6xy+8y2−12x−26y+11=06xy+8y^{2}-12x-26y+11=0.

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Problem 299

Reduce the equation of the following line to its simplest form: 7x2+16xy−23y2−14x−16y−218=07x^{2}+16xy-23y^{2}-14x-16y-218=0.

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Problem 300

Reduce the equation of the following line to its simplest form: 7x2+24xy+38x+24y+175=07x^{2}+24xy+38x+24y+175=0.

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Problem 301

Reduce the equation of the following line to its simplest form: x2−8xy+7y2+6x−6y+9=0x^{2}-8xy+7y^{2}+6x-6y+9=0.

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Problem 302

Reduce the equation of the following line to its simplest form: 9x2+24xy+16y2−40x+30y=09x^{2}+24xy+16y^{2}-40x+30y=0.

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Problem 303

Reduce the equation of the following line to its simplest form: 2x2−72xy+23y2+68x+26y+28=02x^{2}-72xy+23y^{2}+68x+26y+28=0.

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