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.



BP2 = FP2

(x + a)2 = (x-a)+ (y-0)2

            x2 + 2ax + a2 = x2 – 2ax + a+ 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

`         Solution

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

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