Class11- In miscellaneous exercise of chapter …

Any point on Parabola will have co-ordinates (at2, 2at) read as (a t square, 2 a t) . where t is unique for a point on parabola. It is given that all three vertices of equilateral triangle are lying on parabola y2= 4ax. One of point is O(0,0). Other two points will be at t = t1 --> P(at12, 2at1) , at t = t2 --> Q(at22, 2at2). As all three points are vertices of equilateral triangle then OP = OQ. Solving by distance formula, we will get t12= t22 ( read as t1 square = t2 square) , which mean t1= t2, t1= -t2 or -t1= t2. t1 can't be equal to t2 otherwise triangle can't be formed. It means t1 = - t2 , or - t1= t2 , it mean Y co-ordinates(2at) of other two points will have same magnitude but opposite signs. It means that axis of parabola bisects the side of equilateral triangle opposite to vertex (0,0). Also Distance PQ will be length of side of equilateral triangle.