The Imaginary Part of a Parabola

Parabolas are commonly taught in most high school algebra classes. All parabolas are curves defined by a quadratic equation and are symmetrical, open in one direction, and have a vertex.  Generally, a parabola defined by a quadratic equation in standard form y = ax² + bx + c has a line of symmetry at x = ^-b/_2a, opens upward if a is positive (but downward if a is negative), and has a vertex at (^-b/_2a, ^{-b^2 + 4ac}/_4a).  For example, a parabola defined by the quadratic equation y = x² – 6x + 13 have variables a = 1, b = -6, and c = 13, which means it has a line of symmetry at x = ^-b/_2a = ^--6/_2·1 = 3, opens upward because a is positive, and has a vertex at (^-b/_2a, ^{-b^2 + 4ac}/_4a) = (^--6/_2·1, ^{-(-6)^2 + 4·1·13}/_4·1) = (3, 4). 

Some other points that are on the parabolic curve can be found by using the given equation.  For example, in the equation y = x² – 6x + 13, when x = 0, y = 0² – 6·0 + 13 = 13, so (0, 13) is on the curve, and when x = 1, y = 1² – 6·1 + 13 = 8, so (1, 8) is on the curve.  Similar calculations can be made to show that (2, 5), (3, 4), (4, 5), (5, 8), (6, 13), (and so on) are also on this parabolic curve.

However, we can also find imaginary points on any parabolic curve as well.    In the equation y = x² – 6x + 13, when y = 0, then 0 = x² – 6x + 13, and using the quadratic formula x = ^{–b ± √(b^2 – 4ac)}/_2a with variables a = 1, b = -6, and c = 13, yields x = ^{–-6 ± √((-6)^2 – 4·1·13)}/_2·1 = ^{6 ± √(36 – 52)}/₂ = ^{6 ± √-16}/₂ = ^{6 ± 4i}/₂ = 3 ± 2i, so (3 + 2i, 0) and (3 – 2i, 0) should also be on the curve.  Similarly, when y = 3, then 3 = x² – 6x + 13 or 0 = x² – 6x + 10, and using the quadratic formula x = ^{–b ± √(b^2 – 4ac)}/_2a with variables a = 1, b = -6, and c = 13 – 3 = 10, yields x = ^{–-6 ± √((-6)^2 – 4·1·(13 – 3))}/_2·1 = ^{6 ± √(36 – 40)}/₂ = ^{6 ± √-4}/₂ = ^{6 ± 2i}/₂ = 3 ± i, so (3 + i, 3) and (3 – i, 3) should also be on the curve.  Similar calculations can be made to show that (3 + 3i, -5), (3 – 3i, -5), (3 + 4i, -12), (3 – 4i, -12), (and so on) are also on the parabolic curve.

But how can you graph these imaginary points on a coordinate graph?  In an Argand graph, the imaginary part is graphed on a separate axes.  But since both the x-axis and y-axis are already being used, we will have to add a third dimensional z-axis to represent the imaginary part.  This results in the following graph for y = x² – 6x + 13:

It appears that the imaginary part is the same size as the real part, but reflected horizontally at the vertex and then rotated 90° into the third dimension.

Depicting three-dimensional graphs on a two-dimensional medium is difficult to do, even with the help of technology.  To make this easier to draw, let’s rotate the imaginary part back into two dimensions by letting the x-axis serve a double purpose of defining x-values and imaginary values.  This means that in our y = x² – 6x + 13 example, the coordinate (3 – i, 3) would transform to (3 – 1, 3) = (2, 3); the coordinate (3 + i, 3) would transform to (3 + 1, 3) = (4, 3); the coordinate (3 – 2i, 0) would transform to (3 – 2, 0) = (2, 0); the coordinate (3 + 2i, 0) would transform to (3 + 2, 0) = (5, 0); and so on; resulting in the following graph:

This transformation graphically changed the imaginary part to a real part.  We can denote this algebraically by multiplying square root part of the quadratic formula by i.  In other words, the imaginary part of any quadratic equation y = ax² + bx + c (or 0 = ax² + bx + c – y) can be represented by new x-values such that x = ^{–b ± i√(b^2 – 4a(c – y))}/_2a.  Solving for y results in another quadratic equation:

x = –b ± i√(b^2 – 4a(c – y))/2a	(x-values of transformation)
2ax = –b ± i√(b2 – 4a(c – y))	(multiply by 2a)
2ax + b = ±i√(b2 – 4a(c – y))	(add b)
(2ax + b)2 = -1(b2 – 4a(c – y))	(square both sides)
4a2x2 + 4abx + b2 = -b2 + 4ac – 4ay	(distribute)
4ay + 4a2x2 + 4abx + b2 = -b2 + 4ac	(add 4ay)
4ay + 4a2x2 + 4abx = -b2 + 4ac	(subtract b2)
4ay + 4a2x2 + 4abx = 4ac – b2	(rearrange b2)
4ay + 4a2x2 = -4abx + 4ac – 2b2	(subtract 4abx)
4ay = -4a2x2 – 4abx + 4ac – 2b2	(subtract 4a2x2)
y = -ax2 – bx + c – b^2/2a	(divide by 4a)

This proves that the imaginary part of a parabola is another parabola.  (A similar proof can be used to show that the imaginary part of a hyperbola is an ellipse, and that the imaginary part of an ellipse is a hyperbola.)  It also gives us a fast way to forcibly graph the imaginary part of a parabola on a graphing calculator (or some other technology) that would not normally do so.  For example, to graph the full graph of y = x² – 6x + 13, where a = 1, b = -6, and c = 13, we should also graph the imaginary part at the same time using the imaginary transformation equation y = -ax² – bx + c – ^{b^2}_/2a or y = -1x² – (-6)x + 13 – ^{(-6)^2}_/2·1 or y = -x² + 6x – 5.  Graphing y = x² – 6x + 13 (in bold) and y = -x² + 6x – 5 gives the full graph, both real and imaginary, of the parabola:

A great application for graphing the imaginary part of the parabola is to use it as a visual aid for finding the number and types of solutions to a quadratic equation, a common objective in most high school algebra classes.  To find the number and types of solutions to a quadratic equation, students are traditionally taught to calculate the discriminant, which is the part under the square root of the quadratic formula, namely b² – 4ac.  If the discriminant is negative, the square root will result in an imaginary number, and the quadratic formula will yield two imaginary solutions and no real solutions.  If the discriminant is equal to zero, the square root will also be equal to zero, and the quadratic formula will yield one real solution and no imaginary solutions.  Finally, if the discriminant is positive, the square root will also be positive, and the quadratic formula will yield two real solutions and no imaginary solutions.

Discriminant	Solutions
-	0 real, 2 imaginary
0	1 real, 0 imaginary
+	2 real, 0 imaginary

This approach is rather abstract, but the concept can now be reinforced by graphing both real and imaginary parts of the parabola, and examining which part intersects with the x-axis.  For example, let’s say we were asked to find the number and types of solutions to x² – 6x + 13 = 0.  The discriminant is b² – 4ac = (-6)² – 4·1·13 = -16, which is negative, so it will have 0 real solutions and 2 imaginary solutions.  Graphing y = x² – 6x + 13 (with its imaginary transformation equation y = -ax² – bx + c – ^{b^2}_/2a or y = -1x² – (-6)x + 13 – ^{(-6)^2}_/2·1 or y = -x² + 6x – 5) shows that the real part of the parabola does not cross the x-axis but the imaginary part of the parabola crosses the x-axis twice, visually reinforcing the result that there are 0 real solutions and 2 imaginary solutions. 

Using a different example, let’s say we were asked to find the number and types of solutions to x² – 6x + 9 = 0.  Now the discriminant is b² – 4ac = (-6)² – 4·1·9 = 0, so it will have 1 real solution and 0 imaginary solutions.  Graphing y = x² – 6x + 9 (with its imaginary transformation equation y = -ax² – bx + c – ^{b^2}_/2a or y = -1x² – (-6)x + 9 – ^{(-6)^2}_/2·1 or y = -x² + 6x – 9) shows that the vertex crosses the x-axis exactly once.  Recall that the vertex is real, not imaginary, visually reinforcing the result that there is 1 real solution and 0 imaginary solutions. 

Lastly, let’s say we were asked to find the number and types of solutions to x² – 6x + 5 = 0.  Now the discriminant is b² – 4ac = (-6)² – 4·1·5 = 16, which is positive, so it will have 2 real solutions and 0 imaginary solutions.  Graphing y = x² – 6x + 5 (with its imaginary transformation equation y = -ax² – bx + c – ^{b^2}_/2a or y = -1x² – (-6)x + 5 – ^{(-6)^2}_/2·1 or y = -x² + 6x – 13) shows that the real part of the parabola crosses the x-axis twice but the imaginary part of the parabola does not cross the x-axis, visually reinforcing the result that there are 2 real solutions and 0 imaginary solutions. 

x2 – 6x + 13 = 0 0 real solutions 2 imaginary solutions	x2 – 6x + 9 = 0 1 real solution 0 imaginary solutions	x2 – 6x + 5 = 0 2 real solutions 1 imaginary solution

In summary, a quadratic equation is actually comprised of two curves – one real and one imaginary.  The real curve is the traditional parabolic curve.  The imaginary curve is the same size and shares the same vertex as the real curve, but is reflected horizontally and rotated 90° into the third dimension.  Rotating the imaginary curve back into two dimensions helps make it easier to graph both real and imaginary parts of the quadratic equation, and also allows us to quickly determine the number and type of solutions by examining where it intersects with the x-axis.