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

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let me revise it
ok thank you :)
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)

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Other answers:

is (0,1) the focii?
no it is centre
What is the foci?
(5,1) and (-5,1) are focci i mistakenly wrote it as vertices
now i vertices are
@Muhammad_Nauman_Umair \(a^2=9\), not \(a=9\)
oh :( i need to go to sleep i m again mistaken @sirm3d u r write a should b 3 and b should be 4
right not write again mistaken @sirm3d

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