PARABOLA -4-

Parametric Form of a Parabola

Suppose that the equation of a tangent to the parabola y2 = 4ax. … (i)

is y = mx + c. … (ii)

The abscissa of the points of intersection of (i) and (ii) are given by the equation (mx + c)2 = 4ax. But the condition that the straight line (ii) should touch the parabola is that it should meet the parabola in coincident points hence discriminant should be zero

⇒ (mx – 2a)2 = m2c2 … (iii)

⇒ c = a/m.

Hence, y = mx + a/m is a tangent to the parabola y2 = 4ax, whatever be the value of m.

Equation (mx + c)2 = 4ax now becomes (mx – a/m)2 = 0.

⇒ x = a/m2 and y2 = 4ax

⇒ y = 2a/m.

Thus the point of contact of the tangent y = mx + a/m is (a/m2 ,2a/m).

Taking 1/m = t where t is a parameter, i.e., it varies from point to point. The parabola y2 = 4ax as a parametric form is given by the co-ordinate (at2, 2at) and we refer to it as point ‘t’.


Illustration:

Prove that the area of the triangle inscribed in the parabola y2 = 4ax is a2 (t1 – t2) (t2 – t3) (t3 – t1) where t1, t2 and t3 are the vertices.

Solution:

The three points on the parabola are (at12, 2at1), (at22, 2at2) and (at32, 2at3).