No. of points determining a unique parabola

Question

For a parabola, let Focus: $(a_1,b_1)$ Equation of directrix: $y-mx-c=0$ The equation of parabola is,

$\sqrt{(x-a_1)^2+(y-b_1)^2}= \frac{|y-mx-c|}{\sqrt{1+m^2}}$
There are 4 parameters $m,c,a_1,b_1$. To determine these 4 parameters, we need 4 equations. Therefore, if 4 points are given, the four parameters will be solvable thus determining a unique parabola.

Alternatively, a fixed point(focus) and a equation of the directrix should fix a unique parabola. Because, the focus corresponds to two parameters (x and y coordinates) and the directrix corresponds to two parameters (y=mx+c where m,c are parameters). Therefore 4 parameters will be determined by 4 equations and thus 4 points are needed.

But, various sites say that 4 points corresponds to 2 parabolas in the X-Y plane. Why's that?

Answer 1

Five points determine a conic. Since for eccentricity 1 the parabola case determinant $ ( a b - h^2)$ vanishes, only four constants are required.

The text books may be referring to two parabolas their lines of symmetry parallel to the x-,y-with axes ... not sure without seeing the book referred to.

Taking conic general equation with four constants as

$$ a x^2 + 2 \sqrt { a b } x y + b y^2 + 2 f x + 2 g y + 1 = 0 $$

and solving say for $y$ with respect to $x$ in a quadratic equation you get

$$ y = C_1 + C_2 x \pm \sqrt { C_3 x + C_4} $$

where the constants are recast.

If the above two havles of the same parabola are plotted (contour type plots) you get one parabola of arbitrary inclination of its axis of symmetry joined at its vertical line of tangentcy.

Your logic is essentially correct.