Tangents drawn to parabola  y2  =  4ax at the point A and B intersect at C. If S be the focus of the parabola then SA, SC and SB forms

The points A and B on the parabola y² = 4ax will have coordinates (at1², 2at1) and (at2², 2at2) respectively, where t1 and t2 are parameters.

The equations of the tangents at points A and B can be written using the tangent form of the parabola equation, which is yy1 = 2a(x + x1). So, the equations of the tangents at A and B are:

y(2at1) = 2a(x + at1²) and y(2at2) = 2a(x + at2²)

Solving these two equations will give the coordinates of point C.

The focus S of the parabola y² = 4ax is at point (a, 0).

Now, SA, SC and SB will form a triangle. According to the property of a parabola, the tangents at the ends of a focal chord intersect on the directrix and are equally inclined to the axis. Therefore, triangle SAB is isosceles (SA = SB) and triangle SAC is also isosceles (SA = SC). Hence, SA, SC and SB form an equilateral triangle.