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RenaAsf

  • 2 years ago

x^2/9 - (y-1)^2/16 = 1 name the vertices, and foci

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  1. Muhammad_Nauman_Umair
    • 2 years ago
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    let me revise it

  2. RenaAsf
    • 2 years ago
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    ok thank you :)

  3. Muhammad_Nauman_Umair
    • 2 years ago
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    a=9 b=16 As centre is (0,1) c^2=a^2+b^2 then c=+5and -5 as hyperbola is along x-axis so, vertices are (0+5,1) and(0-5,1) i.e. (5,1),(-5,1)

  4. RenaAsf
    • 2 years ago
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    is (0,1) the focii?

  5. Muhammad_Nauman_Umair
    • 2 years ago
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    no it is centre

  6. RenaAsf
    • 2 years ago
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    What is the foci?

  7. Muhammad_Nauman_Umair
    • 2 years ago
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    (5,1) and (-5,1) are focci i mistakenly wrote it as vertices

  8. Muhammad_Nauman_Umair
    • 2 years ago
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    now i vertices are

  9. Muhammad_Nauman_Umair
    • 2 years ago
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    (9,1),(-9,1)

  10. sirm3d
    • 2 years ago
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    @Muhammad_Nauman_Umair \(a^2=9\), not \(a=9\)

  11. Muhammad_Nauman_Umair
    • 2 years ago
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    oh :( i need to go to sleep i m again mistaken @sirm3d u r write a should b 3 and b should be 4

  12. Muhammad_Nauman_Umair
    • 2 years ago
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    right not write again mistaken @sirm3d

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