Question

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

Answer 1

let me revise it

Answer 2

ok thank you :)

Answer 3

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)

Answer 4

is (0,1) the focii?

Answer 5

no it is centre

Answer 6

What is the foci?

Answer 7

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

Answer 8

now i vertices are

Answer 9

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

Answer 10

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

Answer 11

oh :(

i need to go to sleep
i m again mistaken @sirm3d
u r write
a should b 3
and b should be 4

Answer 12

right not write again mistaken @sirm3d