More on Parabolas

We started out the last post by looking at the shape of a set of points equidistant from the focus point and the x-axis. Let us now take a different approach and examine the shape of a set of points equidstant from a focus point and any horizontal line with y = L_y, which we will call the directrix:

sqrt[(X - P_x)² + (Y - P_y)²] = |Y - L_y|
(X - P_x)² + (Y - P_y)² = Y² - 2L_yY + L_y²
(X - P_x)² + Y² - 2P_yY + P_y² = Y² - 2L_yY + L_y²
(X - P_x)² + P_y² - L_y² = 2(P_y - L_y)Y

Here we do the same special case examination: if P_y = L_y, then a bit of algebra will show us that, as before, x = P_x, and we have a vertical line again. We now proceed under the assumption that the focus does not lie on the directrix, or P_y ≠ L_y:

(X - P_x)² + P_y² - L_y² = 2(P_y - L_y)Y
X² - 2P_xX + P_x² + P_y² - L_y² = 2(P_y - L_y)Y

Our shape is once again a parabola, Y = Ax² + Bx + C, with:

D = 2(P_y - L_y)
A = 1 / D
B = -2P_x / D
C = (P_x² + P_y² - L_y²) / D

As a check on our algebra, we can substitute L_y into the above coefficients, and as expected, they transform into the coefficients from the previous case. Armed with this more general result, we again ask the question: given a parabola Y = Ax² + Bx + C, can we find a focus point P and directrix line y = L_y?

This time, we can. The equation for P_x is relatively straightforward:

B = -2P_x / D
P_x = -DB / 2
P_x = -B / 2A

This answer is the same result as we got for the special case parabola where the directrix was fixed as the x-axis. Its presence here is no surprise to anyone who tries to find the root of the derivative of Y = Ax² + Bx + C.

Solving for P_y and L_y is a bit more involved. We start by rearranging the equation for A, to get an equation for their difference:

A = 1/ 2(P_y - L_y)
(P_y - L_y) = 1 / 2A

We then manipulate the equation for C, and we can find an equation for their sum:

C = (P_x² + P_y² - L_y²) / D
C = P_x² / D + (P_y² - L_y²) / D
C = B² / 4A + (P_y - L_y)(P_y + L_y) / D
C = B² / 4A + (P_y + L_y) / 2
(4AC - B²) / 4A = (P_y + L_y) / 2
(P_y + L_y) = -det / 2A

Now that we know the sum and difference of P_y and L_y, we can find their values with ease:

(Py + Ly) + (Py - Ly) = 2Py -det/2A + 1/2a = 2Py Py = (-det + 1) / 4A

(Py + Ly) - (Py - Ly) = 2Ly -det/2A - 1/2a = 2Ly Ly = (-det - 1) / 4A

As a check of our algebra, we observe that when det = -1, we get back the equations we had in the previous post.

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To summarize: given a horizontal line y = L_y, called the directrix, and a point P not on the directrix, calld the focus, there exists a unique parabola Y = Ax² + Bx + C with the following coefficients:

D = 2(P_y - L_y)
A = 1 / D
B = -2P_x / D
C = (P_x² + P_y² - L_y²) / D

The reverse is also true: any parabola of the form Y = Ax² + Bx + C (with A ≠ 0, otherwise it would just be a straight line) has a unique focus point P and directrix line y = L_y given by:

det = B² - 4AC
P_x = -B / 2A
P_y = (-det + 1) / 4A
L_y = (-det - 1) / 4A

The usefulness of this duality is that it works even if the directrix is not horizontal: if you take a directrix line and a focus point P in the 2D plane, they will define a unique parabola, regardless of the orientation of the line. The equation of that parabola is no longer Y = Ax² + Bx + C, but a more complex expression which we will not dwell on anymore here.