Diagonal forms in 2 dimensions are called conic sections, for reasons that will be explained later. They take the form:
ax2 + by2 = c
If b = 0 the solution is the y axis; if a = 0 the solution is the x axis. If both a and b are zero we don't really have a quadratic form. At least one term must have degree 2.
If a and b are both positive, or both negative, the only solution is the origin.
If a and b have opposite sign, replace the coefficients with their square roots and rewrite the equation as follows.
a2x2 - b2y2 = 0
(ax+by) × (ax-by) = 0
The solution is a pair of lines that intersect at the origin and have opposite slope. We've drawn a letter X on the plane.
Divide through by c, so that the constant term is 1. Again, either a or b is nonzero. If b = 0 the solution is empty (a < 0), or two lines parallel to the y axis (a > 0). A similar result holds for a = 0, with lines parallel to the x axis.
If a and b are negative there is no solution.
x2/a2 + y2/b2 = 1
The curve intersects the x axis at ±a, and the y axis at ±b. In fact, we can rescale x by a, and y by b, and obtain the unit circle. Therefore this shape, which is called an ellipse, is a stretched out version of the circle. The curve is multiplied by a in the x direction, and b in the y direction. It is therefore described by the curve:
x(t) ← a×cos(t)
y(t) ← b×sin(t)
Verify that x(t)2/a2 + y(t)2/b2 equals 1, as it should.
Assume, without loss of generality, that a is greater than b. In other words, the ellipse is oriented along the x axis. The major axis (horizontal) has length 2a, and the minor axis (vertical) has length 2b. The semimajor axis (from the origin to the right or left) has length a, and the semiminor axis (from the origin to the top or bottom) has length b.
The ellipse has a focus at the point 0,c on the x axis, where c2 = a2-b2. It has another focus at x = -c. Together these are the two foci (the plural of focus).
My apologies for using the letter c again. At the top of this page, I used c as part of the equation of the ellipse. But we normalized that away, scaling the values of a and b accordingly. From here on in, c is a value that is derived from a and b, according to the formula c2 = a2-b2. It lies on the major axis of the ellipse, which is the x axis in our example. If a = b, a circle, then a2-b2 = 0, and c2 = 0, and the focus is at the origin. If a is much larger than b, than subtracting b2 hardly matters, and c is almost a. The two foci are very close to the left and right edge. So the location of the focus, between the origin and the right edge, is an indicator of the roundness, or flatness, of the ellipse. I'll make this more precise in the next paragraph. Note, it is called a focus because light is focused onto this point by an elliptical mirror. We'll get to that later.
The eccentricity of the ellipse is the ratio c/a. This can range from 0, when the ellipse is a circle, up to (nearly) 1, when the ellipse is very long and skinny.
If p is any point on our ellipse, the distance from p to one focus, plus the distance from p to the other focus, is always 2a. This is clear when p is one of the two points on the x axis, or one of the two points on the y axis. In fact it's true all the way around, as though you had tied two ends of a string (of length 2a) to the foci and traced the curve with a pencil, keeping the string taut.
Let q be the distance to the first focus, and r the distance to the second.
q = sqrt((x-c)2 + y2)
r = sqrt((x+c)2 + y2)
q + r = 2a (start with the conclusion and work backwards)
q = 2a - r
q2 = 4a2 - 4ar + r2
-4cx = 4a2 - 4ar
ar = a2 + cx
a2r2 = a4 + 2a2cx + c2x2
a2×(x2 + 2cx + c2 + y2) = a4 + 2a2cx + c2x2
a2x2 + a2c2 + a2y2 = a4 + c2x2
a2-c2 = b2 (it's how c was defined)
b2x2 + a2c2 + a2y2 = a4
b2x2+ a2y2 = a2b2
x2/a2 + y2/b2 = 1 (formula for the ellipse)
Now that's beautiful! And the converse follows from the same algebra. If two points s and t are 2c units apart, and a is any distance greater than c, draw the points whose distance to s, plus the distance to t, equals 2a. This gives an ellipse, with the formula shown above. The major axis is an extension of the segment st, and the minor axis is perpendicular to st.
The last case to consider is a positive and b negative. (We could have b positive and a negative, but that just reverses the roles of x and y.) replace the coefficients with their inverse square roots and write:
x2/a2 - y2/b2 = 1
Once again the curve intersects the x axis at ±a, but it doesn't intersect the y axis at all. This is obviously not an ellipse. The curve, called a hyperbola, is disconnected, with one branch to the left of the y axis and one branch to the right.
For large x and y, the constant term 1 is almost insignificant. In other words, the hyperbola approaches the degenrate conic x2/a2+y2/b2 = 0. We described this conic at the top of this page. The result was two lines that intersect at the origin, like an infinite letter X. From far away, the two branches of the hyperbola approach the left and right sides of the letter X. The lines of the letter X are called the asymptotes. Therefore the hyperbola approaches its asymptotes.
The noun asymptote has been turned into an adverb, asymptotically, as in, "The curve approaches the line y=4 asymptotically." Often the curve is a hyperbola, but not always. Perhaps it isn't technically correct, but we often write, "The exponential curve approaches the x axis asymptotically as x approaches -infinity." So the more general meaning of asymptote is a line that a curve approaches, steadily and monotonically.
Like an ellipse, the hyperbola has two foci. Set c2 = a2+b2 and plot the points ±c on the x axis. These are farther from the origin than ±a, and they lie inside the two branches of the hyperbola.
The eccentricity is, once again, c/a, but this time the eccentricity is greater than 1.
Like the ellipse, the hyperbola defines, and is defined by, a distance relationship. Let p be a point on the hyperbola, and let q be the distance from p to +c, and let r be the distance from p to -c. On the right branch, r-q = 2a, and on the left branch, q-r = 2a. The algebra is almost identical to that shown above, but let's run through it again, for the left branch.
q - r = 2a (start with the conclusion and work backwards)
q = 2a + r
q2 = 4a2 + 4ar + r2
-4cx = 4a2 + 4ar
-ar = a2 + cx
a2r2 = a4 + 2a2cx + c2x2
a2×(x2 + 2cx + c2 + y2) = a4 + 2a2cx + c2x2
a2x2 + a2c2 + a2y2 = a4 + c2x2
a2-c2 = -b2 (it's how c was defined)
-b2x2 + a2c2 + a2y2 = a4
-b2x2+ a2y2 = -a2b2
b2x2- a2y2 = a2b2
x2/a2 - y2/b2 = 1 (formula for the hyperbola)
Conversely, let s and t be two points 2c units apart, and draw the points whose distance to s, minus the distance to t, equals 2a. This reproduces the hyperbola.
ax2 = y
This shape does not close up like an ellipse, nor does it approach any asymptotes, like a hyperbola. It is the conic in between.
As you might guess, the parabola has a focus. It lies on the y axis, just above the curve. Set c to 1 over 4a, and place the focus at +c on the y axis. Draw a line parallel to the y axis, at y = -c. This is called the directrix.
Like the ellipse and hyperbola, the parabola defines, and is defined by, a distance relationship. A point p is on the parabola iff its distance to the focus equals its distance to the directrix. Let q be the distance to the focus. The distance to the directrix is obviously y+c.
q = y + c (start with the conclusion and work backwards)
q2 = y2 + 2cy + c2
x2 + y2 - 2cy + c2 = y2 + 2cy + c2
x2 = 4cy
ax2 = y (equation for the parabola)
x ← cos(θ)u - sin(θ)v
y ← sin(θ)u + cos(θ)v
Plug these in for x and y, and gather the coefficients for u2, uv, and v2. Label these a′, b′, ad c′, and evaluate b′2-4a′c′. (You probably want a computer to help you do this.) You will find that b2-4ac is multiplied by (sin(θ)2+cos(θ)2)2, which is 1. The discriminant does not change under rotation, hence it can be used to find the type of a conic section, without having to put it into normal form.
A plane can cut through one cone, say the top cone, and create a circle or an ellipse. Now tilt the plane up at a steeper angle, so that it intersects the top cone and runs parallel to the bottom cone. This perfectly balanced conic section is the parabola. Tilt the plane just a bit more, so it intersects both cones, and find the two branches of a hyperbola. That's the idea; the proof is presented in the next section.
The degenerate conic sections also result from a plane and a double cone. Cut across the common apex and find a single point. Let the plane run straight up through both cones, through the common apex, and find two intersecting lines. Finally, a plane can run tangent to the left of the lower cone and the right of the upper cone, producing a single line. The one exception is two parallel lines. A plane and a cone cannot create two parallel lines, even though this is a degenerate conic section. However, a plane and a cylinder can produce two parallel lines, or a single line of tangency, or a circle, or an ellipse. More on this later.