Question

How do you find parametric equation for given curve `y²=4ax`?

Answer 1

`y^2=4ax`

Center `(0, 0)`
Focus `(a, 0)`
Directrix `x=-a)`

Explanation:

The parabola's center is `(0,0)`

It is focus is at point E `(a, 0)`

Its directrix is `x=-a`

Take any point on the parabola. in our case it is F.

The coordinates are `(x, y)`

Then -

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

The Distance between E and F is equal to the distance between F and G.

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

`x^2-2ax+a^2+y^2=x^2+2ax+a^2`

`cancel(x^2)-2ax+cancel(a^2)+y^2=cancel(x^2)+2ax+cancel(a^2)`

`y^2=2ax+2ax`

`y^2=4ax`

Center `(0, 0)`
Focus `(a, 0)`
Directrix `x=-a)`