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Concept 12 / 14Coordinate Geometry
Rotation of Axes and Identifying Conics from the General Second-Degree Equation
Translation and rotation of coordinate axes, the angle tan 2 theta = 2h/(a - b) that removes the xy term, the discriminant h^2 - ab and the determinant Delta for classifying ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 as a pair of lines, circle, parabola, ellipse or hyperbola, and invariants under rotation.
Figure 1
The general second-degree equation
ax2+2hxy+by2+2gx+2fy+c=0
always represents a conic (possibly degenerate). The standard forms studied earlier have their axes along the coordinate axes; a general equation can be brought to standard form by moving and turning the axes.
Translation of axes. Moving the origin to (h0,k0) with parallel axes: x=X+h0, y=Y+k0. Choosing (h0,k0) at the centre of a central conic removes the linear terms. The centre is found by solving ax+hy+g=0 and hx+by+f=0 (the partial derivatives of the left side set to zero).
Rotation of axes. Turning the axes through an angle θ about the origin:
x=Xcosθ−Ysinθ, y=Xsinθ+Ycosθ.
Substituting into ax2+2hxy+by2, the coefficient of XY vanishes when
tan2θ=a−b2h (and θ=4π if a=b).
Rotating by this angle aligns the new axes with the axes of the conic, as the figure shows for x2−xy+y2=3, which becomes 6X2+2Y2=1 after a 45∘ rotation.
Invariants. Under rotation, a+b and ab−h2 do not change, and neither does c (or the determinant Δ below). So after rotation the new coefficients a′,b′ of X2,Y2 are the roots of λ2−(a+b)λ+(ab−h2)=0: they are the eigenvalues of the matrix [ahhb]. This gives the standard form without performing the substitution.
Classification. Let
Δ=ahghbfgfc=abc+2fgh−af2−bg2−ch2.
If Δ=0: a degenerate conic, i.e. a pair of straight lines (real and intersecting if h2>ab, parallel or coincident if h2=ab, or a single point if h2<ab).
If Δ=0:
h2<ab: an ellipse (a circle if also a=b and h=0), or no real locus.
h2=ab: a parabola.
h2>ab: a hyperbola; a rectangular hyperbola if in addition a+b=0.
The quantity h2−ab plays the role of a discriminant, exactly like B2−4AC for the quadratic form. It is invariant under rotation, which is why the type of a conic does not depend on how it is tilted.
Eccentricity from the invariants. For a central conic, once a′,b′ are known the standard form gives e directly. For example, if after translation and rotation the equation is 6X2+2Y2=1, then e=1−62=32.
Conics as focus-directrix loci. A locus defined by 'distance to a point =e times distance to a line' can be written directly: (x−α)2+(y−β)2=e2l2+m2(lx+my+n)2. Expanding gives a general second-degree equation whose type depends only on e (ellipse, parabola, hyperbola for e<1, =1, >1).
Key Takeaways:
- Centre: solve ax+hy+g=0 and hx+by+f=0.
- Rotation removing xy: tan2θ=a−b2h (θ=45∘ if a=b).
- Invariants: a+b, ab−h2, Δ; new X2,Y2 coefficients are the roots of λ2−(a+b)λ+(ab−h2)=0.
- Δ=0: pair of lines. Δ=0: ellipse if h2<ab, parabola if h2=ab, hyperbola if h2>ab.
- Rectangular hyperbola: h2>ab and a+b=0.
- Circle: a=b, h=0.
- Pitfall: reading h as the coefficient of xy instead of half of it.
Brain-Teaser 1: Identify the conic x2+4xy+4y2−3x−6=0.
Solution: a=1, h=2, b=4, g=−23, f=0, c=−6. Then h2−ab=4−4=0. Δ=abc+2fgh−af2−bg2−ch2=−24+0−0−9+24=−9=0. So it is a parabola. Note that the quadratic part (x+2y)2 is a perfect square, which is the characteristic sign of a parabola: its axis is parallel to x+2y=0.
Brain-Teaser 2: Reduce 5x2+4xy+8y2=36 to standard form and find its eccentricity.
Solution: a=5, h=2, b=8, so h2−ab=4−40<0: an ellipse. The new coefficients are roots of λ2−13λ+36=0, i.e. 4 and 9. So the standard form is 4X2+9Y2=36, or 9X2+4Y2=1. Then e=1−94=35. The rotation angle satisfies tan2θ=5−84=−34, but the invariants gave the shape without computing it.
Brain-Teaser 3: For what λ does x2+λxy+y2−5x−7y+6=0 represent a pair of straight lines?
Solution: Here a=b=1, h=2λ, g=−25, f=−27, c=6. Setting Δ=0: abc+2fgh−af2−bg2−ch2=6+2(−27)(−25)2λ−449−425−64λ2=0. That is, 6+435λ−474−23λ2=0; multiply by 4: 24+35λ−74−6λ2=0, so 6λ2−35λ+50=0, giving λ=25 or λ=310. Check λ=25: the quadratic part x2+25xy+y2=(x+2y)(x+2y), and indeed (x+2y−3)(x+2y−2) expands to the given equation.
