Derive the equation of parabola y2=4ax in standard form.
As we have parabola
Take a point P(x,y) on the parabola such that, PF=PB…(i)
By distance formula,
(PF)2=(x−a)2+y2
and (PB)2=(x+a)2
As PF = PB [from (i)],
Now, (x−a)2+y2=(x+a)2
⇒x2−2ax+a2+y2=x2+2ax+a2
Wegety2=4axat[a,0] is the standard form.