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

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

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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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
(9,1),(-9,1)
@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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