Let y 2 = 12x be the parabola and S be its focus. Let PQ be a focal chord of the parabola such that (SP)(SQ) = 147 4 . Let C be the circle described taking PQ as a diameter. If the equation of a circle C is 64x 2 + 64y 2 - \alphax - 64\sqrt3y = \beta , then \beta - \alpha is equal to ________.
Step-by-Step Solution
Key Concept: Apply the core result for parabola parameters and tangents and simplify using the given constraints.
y 2 = 12x a = 3 SP \times SQ = 147 4 (1328) Let P (3t , 6t) and t t = -1 2 1 2 (ends of focal chord) So, Q ( 3 -6 , ) t 2 t S(3, 0) SP \times SQ = PM1 \times QM 2 (dist. from directrix) 3 147 2 = (3 + 3t ) (3 + ) = 2 t 4 2 2 (1 + t ) 49 \Rightarrow = 2 t 12 3 4 2 t = , 4 3 \sqrt3 2 t = $\pm$ ,$\pm$ 2 \sqrt3 -\sqrt3 considering t = 2 9 P( , -3\sqrt3) and Q(4, 4\sqrt3) 4 Hence, diametric circle: 9 (x - 4) (x - ) + (y + 3\sqrt3)(y - 4\sqrt3) = 0 4 25 2 2 \Rightarrow x + y - x - \sqrt3y - 27 = 0 4 \Rightarrow \alpha = 400, \beta = 1728 \beta - \alpha = 1328
Correct Answer: 1328