Let A,B and C be the vertices of a variable right angled triangle inscribed in the parabola y^2=16x. Let the vertex B containing the right angle be (4,8) and the locus of the centroid of ABC be a conic C_o. Then three times the length of latus rectum of C_o is ______
A. A
The equation of the parabola is y^2 = 16x, which gives a = 4. Any point on the parabola can be taken as (4t^2, 8t). The vertex B is given as (4,8), which corresponds to the parameter t = 1. Let the parameters for vertices A and C be t_1 and t_2 respectively. Thus, A (4t_1^2, 8t_1) and C (4t_2^2, 8t_2). The slope of AB is: m_ AB = 8t_1 - 8 4t_1^2 - 4 = 2(t_1 - 1) (t_1 - 1)(t_1 + 1) = 2 t_1 + 1 Similarly, the slope of BC is: m_ BC = 2 t_2 + 1 Since ABC is right-angled at B, AB BC. Therefore, m_ AB m_ BC = -1: 2 t_1 + 1 2 t_2 + 1 = -1 (t_1 + 1)(t_2 + 1) = -4 t_1t_2 + t_1 + t_2 + 5 = 0 Let the centroid of ABC be (h, k). Then: h = 4 + 4t_1^2 + 4t_2^2 3 t_1^2 + t_2^2 = 3h 4 - 1 k = 8 + 8t_1 + 8t_2 3 t_1 + t_2 = 3k 8 - 1 From the perpendicularity condition, we can express t_1t_2 as: t_1t_2 = -5 - (t_1 + t_2) = -5 - ( 3k 8 - 1 ) = -4 - 3k 8 Using the algebraic identity t_1^2 + t_2^2 = (t_1 + t_2)^2 - 2t_1t_2, we substitute the expressions in terms of k: 3h 4 - 1 = ( 3k 8 - 1 )^2 - 2 (-4 - 3k 8 ) 3h 4 - 1 = 9k^2 64 - 3k 4 + 1 + 8 + 3k 4 3h 4 - 1 = 9k^2 64 + 9 3h 4 = 9k^2 64 + 10 3h = 9k^2 16 + 40 9k^2 16 = 3h - 40 k^2 = 16 9 (3h - 40) = 16 3 (h - 40 3 ) Replacing (h, k) with (x, y), the locus C_o is the parabola: y^2 = 16 3 (x - 40 3 ) The length of the latus rectum of this parabola is 16 3 . Three times the length of the latus rectum is 3 16 3 = 16. Answer: 16
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