Equation of Parabola:

Let S is the focus of the parabola, and ZZ₁ is the directrix.

Now, draw SK perpendicular from S on the directrix and bisect SK at A.

So, AS = AK

=> Distance of A from the focus = Distance of A from the directrix

=> A lies on the parabola.

Let SK = 2a

So, AS = SK = 2a

Now, let us take A as the origin, AS as the x-axis and AY a line perpendicular to AS as y-axis.

Then the coordinate of S is (a, 0) and the equation of the directrix ZZ₁  is a = -a

Let P(x, y) be any point on the parabola.

Now, join SP and draw PM and PN perpendicular on the directrix ZZ₁ and x-axis.

Now, from the figure,

PM = NK = AN + NK = x + a

Again since P lies on the parabola,

SP = PM

=> SP² = PM²

=> (x - a)² + (y - 0)² = (x + a)²

=> x² + a² - 2ax + y² = x² + a² + 2ax

=> -2ax + y² =  2ax

=> y² =  2ax + 2ax

=> y² =  4ax

This is the requied equation of the parabola in the standard form

Length of latusrectum:

From the figure,

LSL is the latusrectum of the parabola y² =  4ax

By the symmetry of the curve, SL = SL₁ = λ (say)

So, the coordinate of the L is (a, λ)

Since L lies on the parabola y² =  4ax,

So λ² =  4a*a

=> λ² =  4a²

=> λ = √(4a² )

=> λ = ±2a

Now, LL₁ = 2λ = 2*2a = 4a

So, length of latusrectum is 4a