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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Hyperbola

JEE Main · Mathematics · Hyperbola

The equation& of the chord of the hyperbola 25x^ 2 - 16y^ 2 = 400 ,...

The equation& of the chord of the hyperbola 25x2−16y2=40025x^{2} - 16y^{2} = 400, which is bisected at the point (6,2),( 6, 2 ) , is

Options

  1. A16x−75y=41816x - 75y = 418
  2. B75x−16y=41875x - 16y = 418
  3. C25x−4y=40025x - 4y = 400
  4. DNone of these

Correct answer

B. 75x−16y=41875x - 16y = 418

Step-by-step solution

Given equation of hyperbola is 25x2−16y2=400.25x^{2} - 16y^{2} = 400 .

If (6,2)( 6, 2 ) is the mid-point of the chord, then equation of chord is T=S1.T = {S}_{1} .

⇒25(6x)−16(2y)−400=(25)(36)−16(4)−400⇒ 25\left(6x\right) - 16\left(2y\right) - 400 = \left(25\right)\left(36\right) - 16 ( 4 ) - 400

⇒75x−16y=418⇒ 75x - 16y = 418.

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