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Grade 11Analytical Geometry

find the equation of hyperbola whose centre is origin,transverse axis along x axis,length of conjugate axis is 5 and passing through the point (1,-2) .

Profile image of Sounak Biswas
7 Years agoGrade 11
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1 Answer

Profile image of Arun
ApprovedApproved Tutor Answer7 Years ago
Dear student
 
equation of hyperbola will be
 
x^2/ a^2 – y^2/b^2 = 1
 
also length of conjugate axis = 5 = 2b
 
hence
 
x^2/ a^2 – 4 y^2/25 = 1
 
now
it passes through (1,-2)
 
1 /a^2 – 16/25 = 1
 
1/a^2 = 41/25
 
a^2 = 25/41
 
hence
 
equation of hyperbola is
 
41 x^2/ 25 – 2y^2/25 = 1