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anonymous
 3 years ago
x^2/9  (y1)^2/16 = 1
name the vertices, and foci
anonymous
 3 years ago
x^2/9  (y1)^2/16 = 1 name the vertices, and foci

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Muhammad_Nauman_Umair
 3 years ago
Best ResponseYou've already chosen the best response.1let me revise it

Muhammad_Nauman_Umair
 3 years ago
Best ResponseYou've already chosen the best response.1a=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)

Muhammad_Nauman_Umair
 3 years ago
Best ResponseYou've already chosen the best response.1no it is centre

Muhammad_Nauman_Umair
 3 years ago
Best ResponseYou've already chosen the best response.1(5,1) and (5,1) are focci i mistakenly wrote it as vertices

Muhammad_Nauman_Umair
 3 years ago
Best ResponseYou've already chosen the best response.1now i vertices are

Muhammad_Nauman_Umair
 3 years ago
Best ResponseYou've already chosen the best response.1(9,1),(9,1)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Muhammad_Nauman_Umair \(a^2=9\), not \(a=9\)

Muhammad_Nauman_Umair
 3 years ago
Best ResponseYou've already chosen the best response.1oh :( 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
 3 years ago
Best ResponseYou've already chosen the best response.1right not write again mistaken @sirm3d
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