How do you find the standard form given #9x^2-4y^2-36x-24y-36=0#?

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Answer 1 · 2018-05-03T12:38:02

#( x -2 )^2/ 2^2 - (y +3)^2/3^2 = 1# , equation of a hyperbola .

#9 x^2 -4 y^2 -36 x -24 y -36=0# or

#9( x^2 -4 x ) -4 (y^2 +6 y) = 36# or

#9( x^2 -4 x +4 ) -4 (y^2 +6 y +9) = 36# or

#9( x -2 )^2 - 4 (y +3)^2 = 36# or

#(9( x -2 )^2)/36 - (4 (y +3)^2)/36 = 1# or

#( x -2 )^2/4 - (y +3)^2/9 = 1# or

#( x -2 )^2/ 2^2 - (y +3)^2/3^2 = 1#

This is standard form of the equation of a hyperbola with center

at # ( 2,-3)#

graph{9 x^2-4 y^2-36 x-24 y=36 [-80, 80, -40, 40]} [Ans]