9.4. Rotation of Axes
Thus far, we have only looked at conic sectioins oriented in the standard way, i.e. with vertical and horizontal axes. What about conic sections oriented differently? What do their equations look like? How can we determine what rotation to apply to them to make their axes vertical or horizontal? In this section we will answer both these questions.
The most general equation of a conic section has the following form:
- (9.4.1) Ax2 + Bxy + Cy2 + Dx + Ey + F = 0.
Here, the coefficients A, B, C, D, E, and F are arbitrary real-valued constants with at least one of A, B, or C as well as at least one of D, E, or F nonzero. One should be careful with Equation (9.4.1) however. Although all conic sections can be represented in this way, some values of the constants give rise to figures which are not conic sections. For instance, each of the three degenrate conic sections (a point, a line, or two intersecting lines) may come up, as well as two other cases, namely two parallel lines or the empty set. But most of the time, the resulting figure is a non-degenerate conic section (circle, ellipse, parabola, or hyperbola).
We saw an example of a conic section with nonstandard orientation in Chapter 1, namely the hyperbola y = 1/x. How to we know this is a hyperbola? The first step is to multiply both sides of the equation by x, yielding the equation xy = 1. Now all we have to do is subtract both sides by 1, yielding the equation xy - 1 = 0. Now we see that this equation has the form of (9.4.1) with A = C = D = E = 0, B = 1 and F = -1, so we know that it represents either a conic sections or one of the five degenerate cases. We can definitely rule out the degenerate cases off the bat, since the curve is obviously not any of these. Now of all four non-degenerate conic sections, the only one with two branches is the hyperbola. Thus we know that y = 1/x represents a hyperbola.
How is the hyperbola y = 1/x oriented? Every hyperbola has two axes of symmetry, one passing through the foci and the other perpendicular to the first and meeting it at the center of the hyperbola. Clearly, the line y = x is an axis of symmetry of the hyperbola y = 1/x, better written here as xy = 1, since replacing y with x and x with y does not change this equation. The other axis of symmetry is the line y = -x, since it is the unique line perpendicular to y = x and passing through the origin. The foci of the hyperbola must lie along the line y = x, since this is the only axis which intersects the hyperbola.
So how can we rotate this hyperbola to get it into standard orientation? We want to rotate the line y = x by the correct angle to make it horizontal. In Section 8.5, we studied rotations. In particular, we saw how rotations transform the coordinates of points in the xy-plane. So we choose a point on the line y = x away from the origin; the point (x = 1, y = 1) looks like the simplest choice. We then apply a counterclockwise rotation by an angle θ to this point. The resulting coordinates are then given by
x' = x cos θ - y sin θ = cos θ - sin θ
y' = x sin θ + y cos θ = sin θ + cos θ
Now we want y' to be equal to zero in order for the x'-axis to be horizontal. This means we need to solve the equation sin θ + cos θ = 0. Recall from Section 5.5 how to solve such an equation. The trick with this one is to multiply both sides by cos π/4 = sin π/4 = √2/2, whence we find sin θ cos π/4 + cos θ sin π/4 = sin(θ + π/4) = 0. This equation implies that θ + π/4 must be a multiple of π. The only values of θ from 0 to 2π which work are θ = 3π/4, corresponding to a 3π/4-radian or 135-degree counterclockwise rotation, or θ = 7π/4, corresponding to a 7π/4-radian or 315-degree counterclockwise rotation, which is equivalent to a π/4-radian or 45-degree clockwise rotation. Now since cos 7π/4 = √2/2 and sin 7π/4 = -√2/2, we see that our coordinate transformation becomes
x' = x cos 7π/4 - y sin 7π/4 = √2/2(x + y)
y' = x sin 7π/4 + y cos 7π/4 = √2/2(-x + y)
To see how the equation xy = 1 transforms, we need to apply the inverse transformation, namely
x = x' cos 7π/4 + y' sin 7π/4 = √2/2(x' - y')
y = -x' sin 7π/4 + y' cos 7π/4 = √2/2(x' + y')
Thus, the equation of our hyperbola becomes [√2/2(x' - y')] [√2/2(x' + y')] = 1/2[(x')2 - (y')2] = (x')2/(√2)2 - (y')2/(√2)2 = 1. Now we see that this form of the equation agrees with Equation (9.2.5) with a = b = √2.
Where are the foci of this hyperbola? They lie on the line y = x a distance c = √a2 + b2 = √2 + 2 = 2 to either side of the origin. Thus, the foci lie at (√2, √2) and (-√2, -√2).
Let us look at Equation (9.4.1) again. The main obstacle toward making the orientation of this conic standard seems to be the xy-term. We would like to find a rotation which would make this term disappear. Thus, we apply the inverse of the coordinate transformation
- (9.4.2a) x' = x cos θ - y sin θ
- (9.4.2b) y' = x sin θ + y cos θ
namely the coordinate transformation
- (9.4.2a) x = x' cos θ + y' sin θ
- (9.4.2b) y = -x' sin θ + y' cos θ
to (9.4.1), obtaining a new equation of the form
- (9.4.3) A'(x')2 + B'x'y' + C'(y')2 + D'x' + E'y' + F' = 0.
Sparing the reader of a lot of tedious algebra, the new coefficients A' - F' are given in terms of the old ones and the rotation angle θ by the following formulas:
- (9.4.4a) A' = (A + C)/2 + [(A - C)/2] cos 2θ - B/2 sin 2θ
- (9.4.4b) B' = (A - C) sin 2θ + B cos 2θ
- (9.4.4c) C' = (A + C)/2 + [(C - A)/2] cos 2θ + B/2 sin 2θ
- (9.4.4d) D' = D cos θ - E sin θ
- (9.4.4e) E' = D sin θ + E cos θ
- (9.4.4f) F' = F
Now we want to find the value(s) of θ such that B' = 0. It is easy to see that B' = 0 implies
- (9.4.5) tan 2θ = B / (C - A)
We must be careful with this equation when solving for θ. If A and C are different, we may solve it for θ, obtaining
- (9.4.6) θ = 1/2 arctan (B / (C - A))
But note that (9.4.5) really has two solutions in 2θ, each differing by π, and thus four solutions in θ, each differing by π/2. These four solutions correspond to the four possible rotations which make the axes of the conic vertical and horizontal. Therefore, we might as well take the solution in the interval [0, π/2). The other thing we must be careful about is the case in which A = C, for then the denominator will be equal to zero, implying that tan 2θ is undefined. But this can only happen if 2θ = π/2, i.e. if θ = π/4.
Example 1: Determine the angle in the interval [0, π/2) that we must rotate the coordinates counterclockwise in order to make the axes of the following conic section vertical and horizontal. Also, determine the equation of this conic section in the transformed coordinate system. What kind of conic section is it?
x2 + 2xy + 3y2 - 1 = 0.
Solution: We have A = 1, B = 2, C = 3, D = E = 0, and F = 1. Equation (9.4.6) tells us we must rotate the conic section counterclockwise by an angle θ = 1/2 arctan (B / (C - A)) = 1/2 arctan 1 = (1/2)(π/4) = π/8. Plugging in this value of θ as well as our values of the coefficients A - F into equations (9.4.4), we find
A' = (1 + 3)/2 + [(1 - 3)/2] cos π/4 - 2/2 sin π/4 = 2 - √2/2 - √2/2 = 2 - √2
B' = 0
C' = (1 + 3)/2 + [(3 - 1)/2] cos π/4 + 2/2 sin π/4 = 2 + √2/2 + √2/2 = 2 + √2
D' = E' = 0; F' = -1.
Thus, our transformed equation for the conic section is
(2 - √2)(x')2 - (2 + √2)(y')2 - 1 = 0.
Note that this equation has the form x2 / a2 + y2 / b2= 1, with
Therefore, this conic is an ellipse.
Example 2: Determine the angle in the interval [0, π/2) that we must rotate the coordinates counterclockwise in order to make the axes of the following conic section vertical and horizontal. Also, determine the equation of this conic section in the transformed coordinate system. What kind of conic section is it?
x2 - 2xy + y2 - x - y = 0.
Solution: We have A = C = 1, B = -2, D = E = -1, and F = 0. Since A = C, we have θ = π/4. From Equations (9.4.4) we find
A' = (1 + 1)/2 + [(1 - 1)/2] cos π/2 + 2/2 sin π/2 = 2 + 0 + 2 = 4
B' = 0
C' = (1 + 1)/2 + [(1 - 1)/2] cos π/2 - 2/2 sin π/2 = 2 - 0 - 2 = 0
D' = -1 cos π/4 + 1 sin π/4 = -√2/2 + √2/2 = 0
E' = -1 sin π/4 - 1 cos π/4 = -√2/2 - √2/2 = -√2
F' = 0
Therefore the equation of the transformed conic is
4(x')2 - y'√2 = 0
Adding y'√2 to both sides and then dividing both sides by 4, we find
(x')2 = (√2/4) y'.
By comparing this with Equation (9.3.4), we recognize the transformed conic as a parabola with vertex at the origin and opening up in the x'-direction, i.e. along the line y = x.
There are two important invariant quantities associated with the rotation of a conic section. The first is the discriminant, namely B2 - 4AC. It is a straightforward but tedious exercise, which we leave to the reader, to show that the discriminant is invariant under the transformations (9.4.4). In other words, we have (B')2 - 4A'C' = B2 - 4AC. The other invariant quantity is the trace, which is given simply by A + C. It is much easier to show that the trace is invariant, i.e. A' + C' = A + C; in fact, we can see this immediately by adding Equations (9.4.4a) and (9.4.4c).
The discriminant is a very useful quantity in that it tells us right off the bat the type of conic section represented by a given equation. It turns out that our conic section is an ellipse if the discriminant is negative, a hyperbola if the discriminant is positive, or a parabola if the discriminant is zero.
Example 3: Determine the type of conic section represented by each of the following equations. (The are all non-degenerate).
- (a) 3x2 + 5xy + 4y2 - 2x - 4y + 27 = 0.
- (b) 3x2 + 5xy - 4y2 - 2x - 4y + 27 = 0.
- (c) 4x2 - 12xy + 9y2 + 3x - 2y - 6 = 0.
- (a) The discriminant of this conic section is equal to 52 - (4)(3)(4) = 25 - 48 = -23, so the conic section is an ellipse.
- (b) The discriminant of this conic section is equal to 52 - (4)(3)(-4) = 25 + 48 = 73, so the conic section is an hyperbola.
- (c) The discriminant of this conic section is equal to (-12)2 - (4)(4)(9) = 144 - 144= 0, so the conic section is an parabola.