SOLUTION: Find the equation of hyperbola whose focip(0,root 10) passing through (2,3)

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Question 825770: Find the equation of hyperbola whose focip(0,root 10) passing through (2,3)
Answer by lwsshak3(11628) About Me  (Show Source):
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Find the equation of hyperbola whose focip(0,root 10) passing through (2,3)
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hyperbola has a vertical transverse axis with center at the origin (0,0)
Its standard form of equation:y%5E2%2Fa%5E2-x%5E2%2Fb%5E2=1
c=√10
c^2=a^2+b^2
10=a^2+b^2
b^2=10-a^2
9%2Fa%5E2-4%2F%2810-a%5E2%29=1
90-9a^2-4a^2=a^2(10-a^2)
90-13a^2=10a^2-a^4
a^4-23a^2+90=0
let u=a^2
u^2-23u+90=0
(u-5)(u-18)=0
u=5=a^2
a^2=5
b^2=10-a^2=10-5
b^2=5
or
u=18=a^2
a^2=18
b^2=10-18=-8 (reject)
..
equation:
y%5E2%2F5-x%5E2%2F5=1