Let $ \mathrm{AB} $ is the focal chord of a parabola and $ \mathrm{D} $ and $ \mathrm{C} $ be the foot of the perpendiculars from $ \mathrm{A} $ and $ \mathrm{B} $ on its directrix respectively. If $ \mathrm{CD}=6 $ units and area of trapezium $ \mathrm{ABCD} $ is $ 36 $ square units, then the length (in units) of the chord $ \mathrm{AB} $ is

Question

Let $ \mathrm{AB} $ is the focal chord of a parabola and $ \mathrm{D} $ and $ \mathrm{C} $ be the foot of the perpendiculars from $ \mathrm{A} $ and $ \mathrm{B} $ on its directrix respectively. If $ \mathrm{CD}=6 $ units and area of trapezium $ \mathrm{ABCD} $ is $ 36 $ square units, then the length (in units) of the chord $ \mathrm{AB} $ is,

MARKS App · Accepted Answer

Let S be focus, A S = A D and B S = B C Area of trapezium = 1 2 A D + B C . 6 = 3 ( A S + B C ) = 3 A B Hence, A B = 12 units

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