# If 5 points are necessary to determine a conic, why are only 3 necessary to determine a parabola?

I've just been reading about how 5 points are necessary and sufficient to determine a conic section in Euclidean geometry (https://en.wikipedia.org/wiki/Five_points_determine_a_conic). But if parabolas are conic sections, and if 3 points are sufficient to determine a parabola (since we can solve the resulting system of equations for the parabola equation's constant, coefficient of $x$, and coefficent of $x^2$), how can it be that 5 points are necessary to determine any conic section?

• 3 points are not sufficient to determine a parabola in any direction/ orientation (if you count rotated parabola, you need more point) – user202729 Apr 10 '16 at 6:36
• A circle is a conic section and is determined by only $3$ points. A straight line is a conic section and is determined by only $2$ points. A point is a conic section . . . – bof Apr 10 '16 at 7:40
• As mentioned, three points don't determine a parabola uniquely, if you allow the parabola's axis to point in any direction. See this related question: "Locus of vertex of parabolas through three points" to see that the family of parabolas through three points is somewhat complicated to characterize. – Blue Apr 10 '16 at 9:07
• @Rasputin What, you dare to question Wikipedia? "The degenerate conic is either: a point, when the plane intersects the cone only at the apex; a straight line, when the plane is tangent to the cone (it contains exactly one generator of the cone); or a pair of intersecting lines (two generators of the cone)." – bof Apr 11 '16 at 1:05
• Three points define a parabola, if you know it's a parabola. Those same three points could make a circle, hyperbola, or ellipse, as well. – Bonnaduck Apr 11 '16 at 1:10

Illustrating a comment to OP's own answer. Shown is an entire family of parabolas through three points, indicating that three points alone do not determine a parabola.

As mentioned in my comment, a conic in the coordinate plane has five characteristics: eccentricity, scale, orientation, $x$-location, $y$-location. Our interest in parabolas specifies a characteristic (eccentricity $=1$); the points account for three more characteristics; and the animation cycles through the aspects of the fifth (here, orientation).

• Hi Blue, animation is so nice. What parabola equation do we have with variable orientation as parameter ? – Narasimham Apr 12 '16 at 15:59
• @Narasimham: I'm working on how to present the derivation, but if we have $A=(0,0)$, $B=(c,0)$, $C=(b \cos A, b\sin A)$, then the parabola through $A$, $B$, $C$ whose axis is parallel to the vector $u = (\cos\theta, \sin\theta)$ has this equation: \begin{align}0 &= x^2 \sin A \sin^2\theta - 2x y \sin A \sin \theta \cos\theta + y^2 \sin A \cos^2\theta \\ &- x c \sin A \sin^2\theta + y \left(\; - b \sin^2(A-\theta) + c \cos A \sin^2\theta \;\right)\end{align} – Blue Apr 12 '16 at 18:33
• Like this one Such images burn into visual memory. – Narasimham Jun 4 '17 at 7:52
• Blue Beautiful! – Peter Szilas Jan 29 at 10:24

EDIT 1:

5 points for a conic. A parabola has $\epsilon=1, B^2 - 4 A C =0$ that reduces number of constants/equations to 4.

$$A x^2+ 2 \sqrt {A\,C} x y+C y^2+D x+E y= 1$$

This can be cast into the form making $y$ subject of parabola equation :

$$y = (a x + b) \pm \sqrt { c x + d }$$

If 3 constants are fixed, a single parameter family of parabolas is possible like in the example graphed :

$$y = 2 x - 5 \pm \sqrt {3 x + 2 t}$$

Only 3 constants are not adequate to determine a general parabola . As you can see from the image below, if you are defining a conic, it requires five points to be known. The conics below all share three points (black), but need two more to define what they are.

If we trying to simply define a parabola, it only takes three points because we know it is a parabola, and it can't possibly be a ellipse or hyperbola -- because we decided we already that we are making a parabola! • Bonnaduck.Very nice. – Peter Szilas Jan 29 at 9:44

Thank you to Mikhail for suggesting that knowing from the beginning that the curve is a parabola allows you to determine the specific kind of parabola it is in less than five points. I believe this is the most satisfactory explanation, because given three points in a plane there is also a unique circle passing through them, since the circle equation is

$(x-a)^2+(y-b)^2 = r^2$,

so knowing three points $(x, y)$ allows us to solve for $a, b$ and $r$.

Since circles are also conic sections, clearly three points are not enough to determine whether the curve is a parabola or a circle.

Some suggested more points are necessary to know the "direction/ orientation" of the parabola. There may be something to this. In fact, I believe the extended equation for parabolas (which may be rotated) is

$Ax^2+Bxy+Cy^2+Dx+Ey= F$,

so it seems that perhaps six points may be necessary to determine some parabolas? In any case, if we have that this is the form of the equation for a curve, we have not even eliminated the possibility that it is a circle until we show that either $B$, $D$, or $E$ $\neq{0}$, so this illustrates the first problem I mentioned: needing only three points was based on the initial assumption that we were dealing with a parabola.

All this raises the question of whether 5 points determine a conic section after you know you are dealing with a conic section in the first place, or before -- i.e., whether more points may be needed to determine that the curve defined by these 5 points is indeed a conic section and not some other thing altogether.

• A conic section in the coordinate plane has five characteristics: eccentricity, scale, orientation, $x$-position, and $y$-position. Correspondingly, five (distinct) points will effectively pin-down each of those characteristics so that there's exactly one conic through those points. (Note: Six points are not needed. There are six unknown coefficients in the generic second-degree equation, but they can't all be zero; this allows you to divide-out by one of those coefficients, making it a "1" and leaving only five unknowns.) (continued) – Blue Apr 11 '16 at 1:58
• A circle in the plane is determined by three points, because the term "circle" assumes up-front two characteristics: eccentricity (zero) and orientation (or, in this case, the lack thereof). Three characteristics are in play, so three points are needed. For your question, "parabola" assumes one characteristic (eccentricity = 1); four characteristics remain in play, so four points are needed to pin one down. (The "three points determine a parabola" thing is a simplified case, usually invoked for parabolas whose axis is assumed to be vertical, thereby providing orientation up-front.) (continued) – Blue Apr 11 '16 at 2:05
• The fact that four points are needed to "determine" a parabola means that, if you only have three points identified, then there's an entire family of parabolas through those points, because one of the characteristics remains in play; if you take that characteristic to be orientation, then you can imagine, for each possible direction, a parabola through your three points whose axis aligns with that direction. The "Locus of vertex" reference I made earlier helps show this; I have another illustration in mind that I'll post if I can find time. – Blue Apr 11 '16 at 2:12
• BTW: Solving for the coefficients in the second-degree polynomial is slightly problematic, since it's not necessarily clear from the get-go which coefficient(s) will be non-zero. A "better" approach in this regard is to use the determinant form shown in Equation (8) of the MathWorld "Conic Section" entry. Note that the determinant reinforces the fact that five points are needed, since exactly five rows of the matrix are filled with values taken from the points' coordinates. – Blue Apr 11 '16 at 2:20
• There are six constants in $A x^2+...=0$ but multiplying them all by any non-zero constant gives an equivalent equation, So there are only $5$ degrees of freedom..... Given $3$ points on a conic and given that the conic is a parabola is more info than just being given $3$ points..... $A x^2+B x y +C Y^2+D x+E y+F=0$ determines a parabola iff $B^2=4 A C$ with $A,B,C$ not all $0$. – DanielWainfleet Apr 11 '16 at 5:22