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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
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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?, a detailed solution 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? has been provided alongside types of 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? theory, EduRev gives you an
ample number of questions to practice 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? tests, examples and also practice JEE tests.