Thank you .I have followed the above method.
Formula : P(t) = s(-t3 + 2t2 – t)P1 + s(-t3 + t2)P2 + (2t3 – 3t2 + 1)P2 + s(t3 – 2t2 + t)P3 + (-2t3 + 3t2)P3 + s(t3 – t2)P4
where P is the point on the curve,P1,P2,P3,P4 are the actual points, s is the tanget and it is inversely proportional to t. and t is the tension. I calculate P(t)x and P(t)y co ordinates for the t varying from 0 to 1.
But these points are approximate. Not exact. I need to get the exact position on the curve.
Can anyone suggest me any methods with which I can find the point on the curve?
Or how the points are drawn using the actual points and control points?