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A hyperbola has focus at origin, its eccentricity is √2 and corresponding directrix is x + y + 1 = 0. The equation(s) of its asymptotes is/are
  • a)
    x + 1 = 0
  • b)
    x - 1 = 0
  • c)
    y + 1 = 0
  • d)
    y - 1 = 0
Correct answer is option 'A,C'. Can you explain this answer?
Most Upvoted Answer
A hyperbola has focus at origin, its eccentricity is √2 and correspon...
Equation of this rectangular hyperbola is:
⇒ 2xy + 2x + 2y + 1 = 0
Equation of the pair of asymptotes is:
2xy + 2x + 2y + 1 + c = 0
Δ = 0 ⇒ c = 1
⇒ xy + x + y + 1 = 0
⇒ (x + 1) (y + 1) = 0 represents asymptotes.
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A hyperbola has focus at origin, its eccentricity is √2 and corresponding directrix is x + y + 1 = 0. The equation(s) of its asymptotes is/area)x + 1 = 0b)x - 1 = 0c)y + 1 = 0d)y - 1 = 0Correct answer is option 'A,C'. Can you explain this answer?
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A hyperbola has focus at origin, its eccentricity is √2 and corresponding directrix is x + y + 1 = 0. The equation(s) of its asymptotes is/area)x + 1 = 0b)x - 1 = 0c)y + 1 = 0d)y - 1 = 0Correct answer is option 'A,C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A hyperbola has focus at origin, its eccentricity is √2 and corresponding directrix is x + y + 1 = 0. The equation(s) of its asymptotes is/area)x + 1 = 0b)x - 1 = 0c)y + 1 = 0d)y - 1 = 0Correct answer is option 'A,C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A hyperbola has focus at origin, its eccentricity is √2 and corresponding directrix is x + y + 1 = 0. The equation(s) of its asymptotes is/area)x + 1 = 0b)x - 1 = 0c)y + 1 = 0d)y - 1 = 0Correct answer is option 'A,C'. Can you explain this answer?.
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