Solution:
Given, (y−1)2=21(x−3)
4p(x−h)=(y−k)2isthestandardequationforaright−leftfacingparabolawithvertexat(h,k),andafocallength∣p∣
Rewriting,
4⋅81(x−3)=(y−1)2
On comparing, (h,k)=(3,1),p=81
i. The coordinates of the vertex = (3,1)
ii.
Parabolaissymmetricaroundthex−axisandsothefocusliesadistancepfromthecenter(3,1)alongthex−axis
(3+p,1)=(3+81,1)=(825,1)
So, the coordinates of the focus=(825,1)
iii.
Parabolaissymmetricaroundthex−axisandsothedirectrixisalineparalleltothey−axis,adistance−pfromthecenter(3,1)x−coordinate
x=3−p=3−81⇒x=823
The equation of the directrix: x=823
iv.
Parabolaisoftheform4p(x−h)=(y−k)2andissymmetricaroundthex−axis
Axisofsymmetryisalineparalleltothex−axiswhichintersectsthevertex:y=1
The equation of the axis of symmetry: y=1
v. The ends of the latus rectum: x=825
vi.
The parametric equations of the parabola (y−k)2=4p(x−h) arex=h+pt2 and y=k+2pt.
(h,k)=(3,1),p=81
So, x=3+81t2,y=1+2(81)t
The parametric equations of the parabola: x=3+81t2,y=1+41t
vii. Graph:
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