The parabola is a locus of points, equidistant from a given point, called the Focus and from a given line called the Directrix.
(Length of directrix from V, (AV) = Length of Focus from V ,(FV))
The line AB, a distance ofa, from the y axis is called the Directrix. The line AF is called the axis of symmetry.
BP = FP
BP2 = FP2
(x + a)2 = (x-a)2 + (y-0)2
x2 + 2ax + a2 = x2 – 2ax + a2 + y2
4ax = y2
thus, y2 = 4ax is the equation of the parabola.
The line RQ which is perpendicular to AF is called the latusrectum, V is called the vertex and F the focus of theparabola.
If the vertex of the parabola is translated to a point \(x1,y1), the equation of the parabola becomes
(y-y1)2 = 4a(x-x1)2.
The above equation is said to be in the standard or canonicalform
1. find the focus and directrix of the parabola y2 = 16x
2. write down the equation of the parabola y2– 4y-12x+40 = 0 in its canonical form and hence find i) the vertex; ii) the focus; iii) the directrix of the parabola
1. comparey2 = 16x with y2 = 4ax,
4a = 16 , a = 4
Thus the focus is (4,0) while the directrix is x = -4
2. y2– 4y-12x + 40 = 0
y2– 4y-+4-12x+40 = 0+4 …. (completing the square)
y2– 4y+4 = 12x – 36 …… (rearranging)
(y – 2)2 = 12(x-3) …… (factorising)
But (y – y1)2 = 12(x-x1)
i) hence vertex (x1,y1) = (3,2)
ii) since 4a = 12, a = 3 then the focus (x1+a,y1) = (3+3,2) = (6,2)
iii)the equation of the directrix is x = 3-3 ie x=0
note that the thedirectrix is of equal but opposite distance from the vertex with thefocus
this means the distance between the focus and the vertex = the distance between the directrixand the vertex