Deriving the general equation for a parabola.

Question

I fail to derive the parabola equation from here, which states that the general equation of any conic section is

$\tag{0} ax^2+by^2+cxy+dx+ey+f=0$

where for a parabola we have $ab-c^2=0$ [2].

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, $P=(p,q)$ 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:

$\sqrt{(x-f)^2 + (y-g)^2} = |a(x-p) + b(y-q)| \implies$ $(x-f)^2 + (y-g)^2 = (a(x-p) + b(y-q))^2 \implies$ $\tag{2} x^2 (a^2-1) + y^2(b^2-1) + 2abxy + k_1x + k_2y + k_3 = 0$

Where $k_1, k_2, k_3$ are some constants, which are not important for this discussion.

If we take an example parabola: $x^{2}-4xy+4y^{2}+40x+20y=100$, it would not fit into the equation $2$, but it also does not satisfy the initial condition of $1\cdot4 - 4^2=0$

What I'm doing wrong here?

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[2]: The correct condition is $4ab-c^2=0$, but I left the original text of the question for posterity, even though after reading the answers here I realized that the referenced answer was wrong.

Answer 1

As I wrote in the comments, $x^2-4xy+4y^2+40x+20y=100$ is a parabola and there is no problem here, since the correct condition is $\color{red}4ab=c^2$.

On the other hand, note that\begin{multline}x^2-4xy+4y^2+40x+20y-100=\\=-5\left(\left(\left(\frac2{\sqrt5}\right)^2-1\right)x^2+2\frac2{\sqrt5}\frac1{\sqrt5}xy+\left(\left(\frac1{\sqrt5}\right)^2-1\right)y^2-8x-4y+20\right).\end{multline}Since $\left(\frac2{\sqrt5}\right)^2+\left(\frac1{\sqrt5}\right)^2=1$, this is indeed a parabola. You did not take into consideration the possibility of multiplying the expression of a parabola by a constant.

Answer 2

SOME HINTS:

We are discussing a situation where the parabola axis of symmetry is not parallel to either of the two coordinate axes.

$$ax^2+by^2+cxy+dx+ey+f=0 \tag1 $$

where its discriminant $ 4ab-c^2=0 $ of the quadratic vanishes $ \tag 2 $

Plug in

$$ax^2+by^2\pm 2 \sqrt{ab}+dx+ey+f=0 \tag3 $$

for all values of constant coefficients. What do you call these types of equations?

Can you see two parabola segments represented here?

There are 5 unknown constants; any conic is determined if 5 points are given. If you divide throughout by $f,$ remaining 4 constants suffice for a solution.

Solving a quadratic the two equation can be also cast into form:

$$ y =kx+ly\pm \sqrt{mx+ny} \tag 4 $$

When the quantity under radical sign vanishes, there is common tangent point parallel to y-axis for the two parabolas.

Answer 3

Disclaimer- The notations used are slightly different.

Firstly, the locus of a point which moves on a plane such that the ratio of its distance from a fixed point to its perpendicular distance from a fixed straight line is a conic. The fixed point is the focus and the fixed line is called the directrix of the conic. This ratio is called the eccentricity $e$ and for a parabola $e=1$.The general equation of a conic as stated;

$$ax^2+by^2+2hxy+2gx+2fy+c=0$$

A parabola can be defined when;

(1)$$\frac{SP}{PM}=1$$

(repeating what is said above), Here $SP$ represents the distance of the focus $S(\alpha,\beta)$ of the conic and any point $P(x,y)$ and $PM$ represents the perpendicular distance of the same point $P(x,y)$ from a fixed line known as the directrix.

(2)

When the determinant of the coefficients, $\Delta=\begin{equation*} \begin{pmatrix} a & h & g \\ h & b & f \\ g & f & c \\ \end{pmatrix} \end{equation*}≠0$,

(3)

For a parabola, $h^2=ab$

Now using your given example, $x^2+4y^2-4xy+40x+20y-100=0$;

on comparing gives the values; $a=1$, $b=4$, $h=-2$, $g=20$, $f=10$ and $c=-100$.

Substituting in condition; $h^2=ab$, we get 4=4, hence proving, the conic is a parabola.

Continuing (1),

Refer the below image(apologies for including an image) for the better understanding, The focus is $S(\alpha,\beta)$, the point $P(x,y)$ and the fixed line aka the Directrix is $a'x+b'y+c'=0$, where a',b' and c' are constants; (not to be confused with a,b and c of the general equation of a conic).

Now for a parabola as we saw in (1), $\frac{SP}{PM}=e=1$, substitute SP as $\left|\sqrt{(x-\alpha)^2+(y-\beta)^2)}\right|$ and PM as $\left|\frac{a'x+b'y+c}{\sqrt{(a')^2+(b')^2}}\right|$, If on verifying with your given example by rearranging into this format to get e=1, we can conclude it is a parabola.

Given; $x^2+4y^2-3xy+40x+20y-100=0$, we can rewrite it as the focus-directrix form as mentioned above;

$$\sqrt {x^2+y^2}=\frac{(2x+y-10)}{\sqrt5}$$

we can observe, $e=1$ (as no extra constant is present that supports this equality) and this concludes your example is a parabola with focus $(0,0)$, the directrix is $2x+y=10$, and vertex is $(2,1)$

For last line defense, The given example can be concluded as a parabola when $\Delta ≠ 0$ (as seen in (2) ), which you can easily verify (carefully compare the coefficients).

Answer 4

Another form to handle the parabola.

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

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