Question

Derive the equation of the parabola in the form y^2=4ax

Answer 1

solution :

let’s say you want to use the point (a,b) as your focus and the line y=c as your directrix.

Any point (x,y) on the parabola must be the same distance from (a,b) as the minimum distance from y=c.

Mathematically this means:

sqrt((x-a)^2+(y-b)^2)=y-c

Let’s solve for y:

(x-a)^2+(y-b)^2=(y-c)^2

expand the terms containing y:
(x-a)^2+y^2–2yb+b^2=y^2–2yc+c^2

(x-a)^2+b^2-c^2=2y(b-c)

Obviously this sucks so let’s take a moment here to simplify things.

Note that the point halfway between (a,b) and (a,c) is clearly on the parabola.

This is the vertex. So let’s choose the vertex to be the origin of our coordinate system.

This means that a=0 and b=-c.

Putting this back into our mess above we get:

x^2=4by