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I operate with the parabola definition $\tag{1} \|X-F\| = distance(X, L)$, where $F=(f, g)$ is the focus point, $X=(x, y)$ is a point on the parabola, $L = \{Y | (Y-P)N = 0 \}$ is the directrix line, $N = (a, b)$ is unit-vector normal to the directrix (i.e. $a^2+b^2=1$), $P=(p,q)$ is a point on the directrix.

$distance(X,L) = |(X-P)N| = |(x-p, y-q)(a,b)| = |a(x-p) + b(y-q)|$

If I plug that back into $1$, I get:

$\sqrt{(x-f)^2 + (y-g)^2} = |a(x-p) + b(y-q)| \implies$ $\left(x-f\right)^2 + \left(y-g\right)^2 = \left( a \left( x-p \right) + b \left( y-q \right) \right)^2 \implies$ $\tag{2} k_{x^2}x^2 + k_{y^2}y^2 + k_{xy}xy + k_{x}x + k_{y}y + k = 0$

Where $$ k_{x^2} = a^2-1 ;\\ k_{y^2} = b^2-1 ;\\ k_{xy} = 2ab ;\\ k_{x} = 2(f - a^2p - abq) ;\\ k_{y} = 2(g - abp - b^2q) ;\\ k = a^2p^2+2abpq+b^2q^2-f^2+g^2 $$

How can I derive necessary conditions for the constant coefficients, in order for the equation to be for a parabola?

I suppose we need to prove that not at least one of $k_x, k_y$ should be $\ne 0$, but I don't know how to prove that (by using the parabola definition in equation $1$ above).

We know that $N \ne (0, 0)$, so not both $a, b$ can be $0$. However, even proving at least one of $P, F$ has to be non-0 also is out of my reach for now. When $P=F=0$, I get $(a^2-1)x^2 - y^2 + 2a\sqrt {1-a^2} xy = 0$, and I'm stuck in proving that's not a parabola.

Thanks!

Context

Starting with a wrong premise that the sufficient condition for a parabola is $ab=c^2$, I started the question for the first part of the generic parabolic equation here.

Now, thanks to getting an answer to that question, I understood that that's neither sufficient nor even correct condition, and since the recommendation is to start a new question, that's what I'm doing here, with the wish to get to the generic parabolic equation, together with the sufficient conditions.

Martin Sleziak
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S11n
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1 Answers1

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Given the parabola directrix

$$ p = p_0 + \lambda \vec v,\ \ \ \|\vec v\|=1 $$

and the focus $f$, with $q = (x,y)'$, the parabola has the property:

$$ \|f-q\|^2=\|q-p_0-((q-p_0)\cdot\vec v)\vec v\|^2 $$

so developing and grouping terms we arrive at

$$ q'\cdot(\vec v\cdot\vec v')\cdot q-2(f+(\vec v\cdot\vec v'-I_2)\cdot p_0)\cdot q+\|f\|^2-p_0'\cdot (\vec v\cdot\vec v'-I_2)\cdot p_0=0 $$

here

$$ \vec v\cdot\vec v'=\left(\matrix{v_x^2&v_xv_y\\ v_xv_y & v_y^2}\right)\\ I_2=\left(\matrix{1&0\\ 0 & 1}\right) $$

Now given a quadratic $q'\cdot M\cdot q+B\cdot q + C=0$ it is a parabola if $M$ can be written in the form

$$ M = \left(\matrix{a^2& a b\\ a b& b^2}\right),\ \ \ a^2+b^2\ne 0 $$

and also

$$ (f-p_0)\times\vec v\ne \vec 0 $$

NOTE

The involved parameters are

$$ \cases{ p_0\ \ \, (2)\\ f\ \ \ \ (2)\\ \vec v\ \ \ \ (2) } $$

and the conditions are

$$ \cases{ \|\vec v\| \ne 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)\\ (f-p_0)\times\vec v \ne 0 \ \ (1) } $$

so we have $(6)-(2)=(4)$ characteristic parameters. Now regarding the form

$$ a x^2+b x y + c y^2+d x+ e y + g=0 $$

we have $(6)$ free parameters and $(2)$ conditions to establish a parabola, remaining $(4)$ characteristic parameters.

Cesareo
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  • Thank you for the answer! However, Tom still hasn't introduced matrices nor transposition operation, so I cannot follow your solution yet...is it possible to use the parabolic definition I provided in the question and not use matrices in order to arrive at the general parabola equation? – S11n Jul 08 '23 at 06:36
  • Here we use mainly vector notation. Scalar product, sum of vectors. Matrix notation was used as a short hand notation to avoid large expressions. Perhaps it is interesting to introduce in the formulation of the proposed problem that must be solved with plain algebra. – Cesareo Jul 08 '23 at 11:54
  • Would according to your answer $^2−4+4^2=0 $ also be a parabola? – S11n Jul 08 '23 at 13:23
  • $x^2-4x y + 4y^2=(x-2y)(x-2y)=0$ so we have the product of two lines. To have a parabola we need the contribution of $f$ the focus also. – Cesareo Jul 08 '23 at 13:53
  • I see you added $B \ne 0$...but how can we know that's a sufficient condition, and we're not missing something else? – S11n Jul 08 '23 at 14:50
  • Well. To have a parabola we need a line (directrix) and a focus. Obviously, the line can't contain the focus so concerning the model I used, we have $\vec v, \ p_0$ (directrix) and $f$ (focus) and also that $f-p_0\ne \lambda \vec v$ with $\vec v\ne \vec 0$. – Cesareo Jul 08 '23 at 15:11
  • I don't find that obvious. Let's call the set of conditions $C$ and the statement that a line is parabola $P$. We want to prove $C \iff P$. Therefore, we should prove that without those conditions we get a line which is not a parabola ($P \implies C$), and with those conditions we get a parabola ($C \implies P$). Do you agree? – S11n Jul 11 '23 at 07:02
  • Please. See attached note. – Cesareo Jul 11 '23 at 07:32
  • The added note seems to just link your equation to the one in my question, but I still don't see the proof for why the conditions are sufficient. Sorry if this is painfully obvious to you, but I'm just starting with this, and I try to be a bit more rigorous than an advanced student would do (e.g. they would have confidence to take something as obvious, but I try not to do that at the beginning, to make sure I develop the rigor and deeper understanding of the logic behind the proofs). – S11n Jul 11 '23 at 08:00
  • In time. $(a^2-1)x^2 - y^2 + 2a\sqrt {1-a^2} x y = 0$ cannot be a parabola because if $A^2=1-a^2$ and $B^2=1$ we have $AB\ne a\sqrt{1-a^2}$. – Cesareo Jul 11 '23 at 08:40
  • Instead of patching the answer in comments, can you please make your answer complete and consistent? – S11n Jul 11 '23 at 10:38
  • No more to add. – Cesareo Jul 11 '23 at 10:44