RenaAsf
x^2/9  (y1)^2/16 = 1
name the vertices, and foci



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RenaAsf
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ok thank you :)

Muhammad_Nauman_Umair
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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 xaxis
so,
vertices are (0+5,1) and(05,1) i.e.
(5,1),(5,1)

RenaAsf
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is (0,1) the focii?


RenaAsf
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What is the foci?

Muhammad_Nauman_Umair
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(5,1) and (5,1) are focci i mistakenly wrote it as vertices



sirm3d
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@Muhammad_Nauman_Umair \(a^2=9\), not \(a=9\)

Muhammad_Nauman_Umair
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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

Muhammad_Nauman_Umair
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right not write again mistaken @sirm3d