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

Let S(h, k) be the focus 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)^{2} + (y − k)^{2} = (lx + my + n)^{2} / (l^{2} + m^{2})

⇒ (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 form 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.

ax^{2} +2hxy +by^{2} + 2gx + 2fy +c = 0 represents a parabola if

Δ ≠ 0 and h^{2} = ab ,

where , $ \large \Delta = \left| \begin{array}{ccc} a & h & g \\ h & b & f \\ g & f & c \end{array} \right| $

Special case:

Let the vertex be (α, β) and the axis 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 = Ay^{2} + 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 parabola is y = A’x^{2} + B’x + C’

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

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

$\large \sqrt{(x-3)^2 + (y-4)^2} = |\frac{x-y+5}{\sqrt{1+1}}|$

⇒ (x – 3)^{2} + (y + 4)^{2} = (x − y + 5)^{2}/2

⇒ x^{2} + y_{}^{2} + 2xy – 22x + 26y + 25 = 0

⇒ (x + y)^{2} = 22x – 26y – 25.

Illustration : Two mutually perpendicular chords OA and OB are drawn through the vertex ‘O’ of a parabola y^{2} = 4ax. Then find the locus of the circumcentre of triangle OAB.

Solution:

Let (-g, -f) be the circumcentre of DOAB.

Since OA is perpendicular to OB

⇒ t_{1}t_{2} = – 4 ….(1)

Clearly (-g, -f) is the mid-point of AB.

⇒ – g = a(t_{1}^{2} + t_{2}^{2})/2 and – f = (t_{1} + t_{2}) ….(2)

From (1) and (2), we get

(t_{1} + t_{2})^{2} = t_{1}^{2} + t_{2}^{2} + 2t_{1}t_{2}

⇒ f ^{2 }/a^{2 } = −2 g/a − 8

⇒ required locus is y^{2} = 2ax – 8a^{2}.

Illustration : Find the equation of the parabola whose focus is (–6, –6) and vertex (-2, 2).

Solution:

Let S(–6, –6) be the focus and A(–2, 2) vertex of the parabola. On SA take a point K (x_{1}, y_{1}) such that SA = AK. Draw KM perpendicular on SK. Then KM is the directrix of the parabola.

Since A bisects SK ,

$\large (\frac{-6+x_1}{2} , \frac{-6+y_1}{2})$ = (-2 , 2)

⇒ – 6 + x_{1} = – 4, and – 6 + y_{1} = 4 or (x_{1}, y_{1}) = (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

y – 10 = (-1/2)(x-2) (From (1))

⇒ 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. From SP = PM, the equation of the parabola is

$\large \sqrt{(x+6)^2 + (y+6)^2} = \frac{|x+2y-22|}{\sqrt{1^2+2^2}}$

or 5(x^{2} + y^{2} + 12x + 12y + 72) = (x + 2y – 22)^{2}

or 4x^{2} + y^{2} – 4xy + 104x + 148y – 124 = 0

or (2x – y)^{2} + 104x + 148y – 124 = 0

Exercise :

(i) Prove that the equation y^{2} + 2ax + 2by + c = 0 represents a parabola whose axis is parallel to x–axis. Find its vertex and the equation of the double ordinate through the focus.

(ii) Find the equation of the parabola whose vertex is at (2, 1) and the directrix is x = y – 1.

(iii) Show that the semi-latus rectum of a parabola is the harmonic mean between the segments of any focal chord.

(iv) Find the length of the side of an equilateral D inscribed in the parabola y^{2} = 4ax so that one angular point is at the vertex.