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Question: What will be equation of that chord of hyperbola \(25x^{2} - 16y^{2} = 400\), whose mid point is (5,...

What will be equation of that chord of hyperbola 25x2−16y2=40025x^{2} - 16y^{2} = 400, whose mid point is (5, 3)

A

115x−117y=17115x - 117y = 17

B

125x−48y=481125x - 48y = 481

C

127x+33y=341127x + 33y = 341

D

15x+121y=10515x + 121y = 105

Answer

125x−48y=481125x - 48y = 481

Explanation

Solution

According to question, xy=c2/2xy = c^{2}/2Equation of required chord is S1=TS_{1} = T ......(i)

Here S1=25(5)2−16(3)2−400S_{1} = 25(5)^{2} - 16(3)^{2} - 400 = 625−144−400=81625 - 144 - 400 = 81 and T=25xx1−16yy1−400T = 25xx_{1} - 16yy_{1} - 400, where x1=5x_{1} = 5, t=t3t2−t4t′+t′t = t^{3}t^{2} - t^{4}t^{'} + t^{'}

⇒25x(5)−16y(3)−400=125x−48y−40025x(5) - 16y(3) - 400 = 125x - 48y - 400So, from (i)

required chord is 125x−48y−400=81125x - 48y - 400 = 81 ⇒ 125x−48y=481125x - 48y = 481