PARABOLA -5-

THE GENERAL EQUATION OF A PARABOLA

We shall now obtain the equation of a parabola when the focus is any point and the dircectrix is any line.

Let (h, k) be the focus S and lx + my + n = 0 the equation of the directrix ZM of a parabola. Let (x, y) be the coordinates of any point P on the parabola. Then the relation, PS = distance of P from ZM, gives

(x – h+) + (y – k)2 = (lx + my + n)2/(l2 + m2)

⇒ (mx – ly)2 + 2gx + 2fy + d = 0.

This is the general equation of a parabola. It is clear that second-degree terms in the equation of a parabola form a perfect square.

The converse is also true, i.e. if in an equation of the second degree, the second-degree terms from a perfect square then the equation represents a parabola, unless it represents two parallel straight lines.


Note:
The general equation of second degree i.e. ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents a parabola if Δ ≠ 0 and h2 = ab. (Δ = abc + 2fgh – af2 – bg2 – ch2).

Special case:

Let the vertex be (α, β) and the axis to be parallel to the x-axis. Then the equation of parabola is given by (y – β)2 = 4a (x – α) which is equivalent to x = Ay2 + By + C.

If three points are given we can find A, B and C.

Similarly, when the axis is parallel to the y-axis, the equation of the parabola is y = A’x2 + B’x + C’.


Illustration:

Find the equation of the parabola whose focus is (3, –4) and directrix
x – t + 5 = 0.

Solution:

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

√((x-3)2 (y+4) )=x-y+5/√(1+1)

⇒ (x – 3)2 + (y + 4)2 = (x-y+5)2/2

⇒ x2 + y2 + 2xy – 22x + 26y + 25 = 0.

⇒ (x + y)2 = 22x – 26y – 25.


Illustration:

Find the equation of the parabola having focus (–6, 6) and verte (–2, 2).

Solution:

Let S(–6, –6) be the focus and A(–2, 2) the vertex of the parabola. On SA take a point K (x1, y1) such that SA = AK. Draw KM perpendicular on SK. Then KM is the directrix of the parabola.

Since A bisects SK, ((-6+x1)/2,(-6+y1)/2) = (–2, 2)

⇒ – 6 + x1 = – 4, and – 6 + y1 = 4

Or (x1, y1) = (2, 10).

Hence the equation of the directrix KM is y – 10 = m (x + 2). … (1)

Also gradient of

SK = (10-(-6))/(2-(-6) )=16/8 = 2; ⇒ m = (-1)/2

So that equation (1) becomes y – 10 = –1/2 (x – 2)

or x + 2y – 22 = 0 is the directrix.

Next, let PM be a perpendicular on the directrix KM from any point

P(x, y) on the parabola.

Form SP = PM, the equation of the parabola is

√({(x+6)+(y+6)2 } )=(x+2y-22)/((12+22 ) )

or 5 = (x2 + y2 + 12x + 12y + 72) = (x + 2y – 22)2

or 4x2 + y2 – 4xy + 104x + 148 y – 124 = 0.

or (2x – y)2 + 104x + 148y – 124 = 0.


Illustration:

If the point (2, 3) is the locus and x = 2y + 6 is the directrix of a parabola, find

(i) the equation of the axis,

(ii) the co-ordinates of the vertex,

(iii) length of the latus rectum,

(iv) equation of the latus rectum.


Solution:

(i) We know that the axis of a parabola is the line through the focus

And perpendicular to the directrix.

The equation of any line passing through the focus (2, 3) is

y – 3 = m (x – 2) ⇒ mx – y = 3 – 2m

If the line be perpendicular to the directrix x – 2y = 6, we have,

m (1/2) = – 1 ⇒ m = – 2.

Hence the equation of the axis is y – 3 = – 2 (x – 2) ⇒ 2x + y = 7.

(ii) The co-ordinates of the point of intersection (say) A of the directrix x – 2y = 6 and the axis 2x + y = 7 are obtained by solving the two equations; thus they are (4, –1). Since the vertex is the middle point of A (4, –1) and the focus S(2, 3); the co-ordinates of the vertex are ((4+2)/2,(3-1)/2), i.e. (3, 1).

(iii) Since OS = √((3-2)2+(1-3)2 )=√5,

The length of the latus rectum = 4OS = 4√5.

(iv) Since the latus rectum is the line through the focus parallel to the directrix, its equation is x – 2y + c = 0, where c is given by 2 – 2.3 + c = 0, i.e. c = 4.

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