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Equation of the chord of the hyperbola 25x2–16y2= 400 which is bisected at the point (6, 2) isa)16x –75y = 418b)75x –16y = 418c)25x –4y = 400d)None of theseCorrect answer is option 'B'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Equation of the chord of the hyperbola 25x2–16y2= 400 which is bisected at the point (6, 2) isa)16x –75y = 418b)75x –16y = 418c)25x –4y = 400d)None of theseCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Equation of the chord of the hyperbola 25x2–16y2= 400 which is bisected at the point (6, 2) isa)16x –75y = 418b)75x –16y = 418c)25x –4y = 400d)None of theseCorrect answer is option 'B'. Can you explain this answer?.

Solutions for Equation of the chord of the hyperbola 25x2–16y2= 400 which is bisected at the point (6, 2) isa)16x –75y = 418b)75x –16y = 418c)25x –4y = 400d)None of theseCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.

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