# The General Parabola Family

When I was determining the rotation equations of a parabola, I realized that there are 4 parabolas that can be formed with a combination of A, B, and C, while D, E, and F are kept constant. We can change the signs of the first 3 to give us four parabolas. They make two pairs, where each pair is congruent, meaning that have the same a that determines how wide or narrow the parabola is.

The family of parabolas is given by the following equations:

(1) $$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$$

(2) $$-Ax^2 - Bxy - Cy^2 + Dx + Ey + F = 0$$

(3) $$Ax^2 - Bxy + Cy^2 + Dx + Ey + F = 0$$

(4) $$-Ax^2 + Bxy - Cy^2 + Dx + Ey + F = 0$$

Parabolas (1) and (2) are congruent - they have the same shape and size but different orientations. Their angle of rotation are the same but in opposite directions.

Likewise, parabolas (3) and (4) are congruent and their angle of rotation are the same but in opposite directions.

These equations are the only possible parabolas because the sign of A and C have to be the same in order for the determinant $$B^2-4AC$$ to be 0.

The figures below show two examples of the parabola family.

## Common Tangents

Looking at the images, each pair of parabolas have one common tangent point. We will find the common tangent points. We will take the first pair given by equations (1) and (2) by setting them equal to each other.

(i) $$Ax^2 + Bxy + Cy^2 + Dx + Ey + F =$$ $$-Ax^2 - Bxy - Cy^2 + Dx + Ey + F$$

(ii) $$Ax^2 + Bxy + Cy^2 = -Ax^2 - Bxy - Cy^2$$

(iii) $$2Ax^2 + 2Bxy + 2Cy^2 = 0$$

(iv) $$Ax^2 + Bxy + Cy^2 = 0$$

(v) $$x = \dfrac{-By \pm \sqrt{B^2y^2-4ACy^2}}{2A}$$  (Solving the quadratic for x)

(vi) $$x = \dfrac{-By \pm y\sqrt{B^2-4AC}}{2A}$$ (What a surprise!)

(vii) $$x = -\dfrac{B}{2A}y$$

We have only one solution. We need to eliminate x by substituting this into (i) and solving to find the y value based solely on the coefficients.

(viii) $$A\left(\dfrac{-By}{2A}\right)^2 + B\left(\dfrac{-By}{2A}\right)y + Cy^2 + D\left(\dfrac{-By}{2A}\right) + Ey + F =0$$

(viii) $$\dfrac{B^2}{4A}y^2 - \dfrac{B^2}{2A}y^2 + Cy^2 - \dfrac{BD}{2A}y + Ey + F =0$$

(ix) $$\left(\dfrac{-B^2+4AC}{4A}\right)y^2 - \left(\dfrac{BD-2AE}{2A}\right)y + F =0$$  (Another great surprise.)

(x) $$\left(\dfrac{BD-2AE}{2A}\right)y = F$$

(xi) $$y = \dfrac{2AF}{BD-2AE}$$

And since $$x = -\frac{B}{2A}y$$, the x value is: $$-\frac{BF}{BD-2AE}$$.

The common tangent point of the pair is $$\left(-\dfrac{BF}{BD-2AE}, \dfrac{2AF}{BD-2AE}\right)$$.

Likewise, the solution for equations (3) and (4) is $$\left(-\dfrac{BF}{BD+2AE}, -\dfrac{2AF}{BD+2AE}\right)$$.

Both of these points are tangents because we only got 1 solution. If it was an intersection, the parabola has to have another solution. It is simply not possible for two parabolas to intersect at only 1 point.

Notice that if F, the constant term, is 0, then we only have one tangent point. The parabolas are all tangent at the origin! Figure 1 is an example of all four tangnet at the origin.

## Common Intersections

In addition to the common tangents, the parabolas have common intersections also. The solutions to equation pair (1) and (3) is the same as the pair (2) and (4). Their solutions are:

(i) $$Ax^2 + Bxy + Cy^2 + Dx + Ey + F =$$ $$Ax^2 - Bxy + Cy^2 + Dx + Ey + F$$

(ii) $$Bxy = - Bxy$$

(iii) $$xy = 0$$

(iii) $$x = 0$$ and $$y = 0$$

We have two solutions: the x- and y-intercepts of the parabolas are the two solutions. We can find the value of the other coordinate by substitution. If x = 0, then

(iv) $$A\cdot0^2 + B\cdot 0\cdot y + Cy^2 + D\cdot0 + Ey + F = 0$$

(v) $$Cy^2 + Ey + F = 0$$

(vi) $$y = \dfrac{-E \pm \sqrt{E^2-4CF}}{2C}$$

One pair of intersection points are $$\left(0, \frac{-E \pm \sqrt{E^2-4CF}}{2C} \right)$$.

Now, if y = 0, then:

(iv) $$Ax^2 + Bx\cdot 0 + C\cdot 0^2 + Dx + E\cdot 0 + F = 0$$

(v) $$Ax^2 + Dx + F = 0$$

(vi) $$x = \dfrac{-D \pm \sqrt{D^2-4AF}}{2A}$$

The other pair of intersection points are $$\left(\frac{-D \pm \sqrt{D^2-4AF}}{2A}, 0 \right)$$.

Again, if F = 0, then one of the intersection points are actually the tangent points at the origin.

In Figure 2, we have more than 4 intersections points. We actually have 7 because one of them is a tangent point rather than an intersection point. So, we can have 8 points of intersections for a parabola family. We simply need to substitute x = 0 and y = 0 in equations (2) and (4).

The other solutions are $$\left(0, \frac{E \pm \sqrt{E^2+4CF}}{2C} \right)$$ and $$\left(\frac{D \pm \sqrt{D^2+4AF}}{2A}, 0 \right)$$.

Because F = 0, we do not have 8 intersection points. We have 5 instead. When F = 0, the parabolas will cross the origin because (0,0) will be a point on the parabola. That eliminated the points of tangency from 4 points to 1.

Also because F = 0, the 2 intersection points on the x-axis are equidistant from the origin and the 2 intersection points on the y-axis are also equidistant from the origin.

In Figure 4, we have 7 points of intersectin identified. Point T became a point of tangency instead of an intersection point because $$E^2 - 4CF$$ evaluated to 0 in the determinant of one of the intersection point equations.

## Rotation to Standard Orientation

We will rotate the variations of parabola family $$x^2+4xy+4y^2+3x-4y+1=0$$ to standard orientation and examine their graphs to see if they retain the same number of tangencies and intersections.

### Parabola 1

The equation of Parabola 1 is $$x^2+4xy+4y^2+3x-4y+1=0$$. If you have read my other pages, then you know this parabola will rotate in the negative direction and will have an orientation that opens left or right because |C| > |A|. The angle of rotation is $$\frac{1}{2}\cot^{-1}\frac{-3}{4} = -26.57^{\circ}$$.

We will skip some intermediate steps and go right into the rotation equation of a parabola for which |C| > |A|:

(i) $$(A+C)y^2 + \left(\dfrac{D\sqrt{|C|} \pm E\sqrt{|A|}}{\sqrt{|A+C|}}\right)x \text{ } +$$ $$\left(\dfrac{E\sqrt{|C|} \mp D\sqrt{|A|}}{\sqrt{|A+C|}}\right)y + F = 0$$

(ii) $$(1+4)y^2 + \left(\dfrac{3\sqrt{4} - (-4)\sqrt{1}}{\sqrt{5}}\right)x \text{ } +$$ $$\left(\dfrac{-4\sqrt{4} + 3\sqrt{1}}{\sqrt{5}}\right)y + 1 = 0$$

(iii) $$5y^2 + \dfrac{10}{\sqrt{5}}x - \dfrac{5}{\sqrt{5}}y + 1 = 0$$

The equation of our parabola in standard orientation is $$5y^2 + \frac{10}{\sqrt{5}}x - \frac{5}{\sqrt{5}}y + 1 = 0$$ or $$x = -\frac{\sqrt{5}}{2}y^2+\frac{1}{2}y-\frac{\sqrt{5}}{10}$$.

### Parabola 2

The equation of Parabola 2 is $$-x^2-4xy-4y^2+3x-4y+1=0$$. The angle of rotation is the same as Parabola 1. And because |C| > |A|, we will use the same equation as above for rotation.

(i) $$(A+C)y^2 + \left(\dfrac{D\sqrt{|C|} \pm E\sqrt{|A|}}{\sqrt{|A+C|}}\right)x \text{ } +$$ $$\left(\dfrac{E\sqrt{|C|} \mp D\sqrt{|A|}}{\sqrt{|A+C|}}\right)y + F = 0$$

(ii) $$(-1+(-4))y^2 + \left(\dfrac{3\sqrt{4} - (-4)\sqrt{1}}{\sqrt{5}}\right)x \text{ } +$$ $$\left(\dfrac{-4\sqrt{4} + 3\sqrt{1}}{\sqrt{5}}\right)y + 1 = 0$$

(iii) $$-5y^2 + \dfrac{10}{\sqrt{5}}x - \dfrac{5}{\sqrt{5}}y + 1 = 0$$

The equation of our parabola in standard orientation is $$-5y^2 + \frac{10}{\sqrt{5}}x - \frac{5}{\sqrt{5}}y + 1 = 0$$ or $$x = \frac{\sqrt{5}}{2}y^2+\frac{1}{2}y-\frac{\sqrt{5}}{10}$$. The only term that changed is the coefficient of the y² term. The parabola opens right.

Figure 5 show Parabola 1 in blue and Parabola 2 in orange. Both are still tangent at one point. If we rotate them by the same angle measure, we expect the point of tangecy to be retained.

### Parabola 3

The equation of Parabola 3 is $$x^2 - 4xy + 4y^2 + 3x - 4y + 1 = 0$$. The angle of rotation is $$\cot^{-1}\frac{1-4}{-4} = 25.67^{\circ}$$ - the same angle measure but positive this time. We will use the same equation as above for rotation because |C| > |A|.

(i) $$(A+C)y^2 + \left(\dfrac{D\sqrt{|C|} \pm E\sqrt{|A|}}{\sqrt{|A+C|}}\right)x \text{ } +$$ $$\left(\dfrac{E\sqrt{|C|} \mp D\sqrt{|A|}}{\sqrt{|A+C|}}\right)y + F = 0$$

(ii) $$(1+4)y^2 + \left(\dfrac{3\sqrt{4} + (-4)\sqrt{1}}{\sqrt{5}}\right)x \text{ } +$$ $$\left(\dfrac{-4\sqrt{4} - 3\sqrt{1}}{\sqrt{5}}\right)y + 1 = 0$$

(iii) $$5y^2 + \dfrac{2}{\sqrt{5}}x - \dfrac{11}{\sqrt{5}}y + 1 = 0$$

The equation of our parabola in standard orientation is $$5y^2 + \frac{2}{\sqrt{5}}x - \frac{11}{\sqrt{5}}y + 1 = 0$$ or $$x = -\frac{5\sqrt{5}}{2}y^2+\frac{11}{2}y-\frac{\sqrt{5}}{2}$$.

The width of this parabola changed because $$a = -\frac{5\sqrt{5}}{2}$$. We expect Parabol 4 to have the same a but a negative sign.

### Parabola 4

The equation of Parabola 4 is $$-x^2 + 4xy - 4y^2 + 3x - 4y + 1 = 0$$. The angle of rotation is $$\cot^{-1}\frac{1-4}{-4} = 25.67^{\circ}$$ - the same angle measure but positive this time. We will use the same equation as above for rotation because |C| > |A|.

(i) $$(A+C)y^2 + \left(\dfrac{D\sqrt{|C|} \pm E\sqrt{|A|}}{\sqrt{|A+C|}}\right)x \text{ } +$$ $$\left(\dfrac{E\sqrt{|C|} \mp D\sqrt{|A|}}{\sqrt{|A+C|}}\right)y + F = 0$$

(ii) $$(-1+(-4))y^2 + \left(\dfrac{3\sqrt{4} + (-4)\sqrt{1}}{\sqrt{5}}\right)x \text{ } +$$ $$\left(\dfrac{-4\sqrt{4} - 3\sqrt{1}}{\sqrt{5}}\right)y + 1 = 0$$

(iii) $$-5y^2 + \dfrac{2}{\sqrt{5}}x - \dfrac{11}{\sqrt{5}}y + 1 = 0$$

The equation of our parabola in standard orientation is $$-5y^2 + \frac{2}{\sqrt{5}}x - \frac{11}{\sqrt{5}}y + 1 = 0$$ or $$x = \frac{5\sqrt{5}}{2}y^2+\frac{11}{2}y-\frac{\sqrt{5}}{2}$$.

The width of this parabola changed because $$a = \frac{5\sqrt{5}}{2}$$.