What's the correct way to derive the eq. of a parabola without knowing the conventional values of focus and directrix?

Question

According to the convention that we are taught in our schools, focus = (a,0) [for a standard parabola] and directrix: x=-a. But is there any way to obtain the equation without knowing these conventions?

I know the basic definition is this: A Parabola is the locus of all points whose distance from a point is equal to their perpendicular distance from a line known as the directrix.

But is it possible to derive the equation from the ground up using just the definition and not making any of the assumptions like I did in the first line?

P.S - If you have the link to the derivation or any article that is related to this, please do send it!

Thankyou so much!

Answer 1

EDIT: This got a downvote, which made me check to see what was off, there's an addition at the asterisk below.

If you did not know the focus or directrix, but you DO have a curve in a coordinate space, and you know it is a parabola, there are several options. I presume the axis is vertical, with upward opening.

1. Find the vertex $(p,q)$. Then $y+q=(x-p)^2$* , if this is a $y=x^2$ form of a parabola. If it is wider or narrower, you need to determine a scalar, $m$ to account for that, most easily done by examining the value of the curve at $p+1$, to yield $y+q=m(x-p)^2$. This is the only simple option if you have imaginary roots.

2. Identify where the parabola touches/crosses zero, these are the roots, and there are at most two distinct real roots. If one, $y=(x-r_1)^2$. If two, $y=(x-r_1)(x-r_2)$.

3. One or both of those methods will work for any parabola, and if it opens downward, just add a negative sign to the RHS. If you then wish to find the geometric focus and directrix, reverse the process below.

Start with a parabola with axis on the y-axis and vertex at the origin:

Then for any point $(x,y)$ on the parabola:

Since the distances from $(x,y)$ to the focus and directrix are the same, use Euclidean distance (i.e., Pythagorean theorem), but square both sides of the equality so you never deal with square roots

$\sqrt{(x-0)^2+(y-a)^2}=\sqrt{(x-x)^2+(y-(-a))^2}$

$(x-0)^2+(y-a)^2=(x-x)^2+(y-(-a))^2$

$(x-0)^2+(y-a)^2=(y+a)^2$

$x^2+y^2-2ay+a^2=y^2+2ay+a^2$

$x^2-2ay=2ay$

$x^2=4ay$

P.S. You will find this brief derivation in virtually any book on coordinate geometry, also in many calculus texts since parabolas are some of the simplest curves to examine.

Answer 2

Parabola by definition is the locus of points equidistant from a given point and a given line.

To derive its equation assume a general line of the form ax+by+c=0 and a general point (h,k). Use the standard formulas for distance from a line and distance from a point and equate them.