SOLUTION: Dear Sir/Ma'am Please help me solve this problem. Given the equation 36x2-36y2=1296, Sketch and determine the parts of the hyperbola. Parts of the Hyperbola 1. Center 2.

Question 1181019: Dear Sir/Ma'am
Please help me solve this problem.
Given the equation 36x2-36y2=1296, Sketch and determine the parts of the hyperbola.
Parts of the Hyperbola
1. Center
2. Foci
3. Vertices
4. Length of Conjugate Axis B1, B2
5. End Points of Latus Rectum E1, E2, E3, E4
6. Asymptotes
7. Eccentricity
8. Length of LR
9. Length of Conjugate Axis
10. Length of Traverse Axis
Thank you so much and more power.
God bless you
Lorna

Answer by MathLover1(20850) (Show Source): You can put this solution on YOUR website!
Given the equation
}
Sketch and determine the parts of the hyperbola.
} .......write in standard form , both sides divide by
}
}
=> and
=> =>
=> =>
=>=>
Parts of the Hyperbola
1. Center: (,)
2. Foci: is or

(,) and (,)
3. Vertices: one vertex is at (, ), and the other is at (, )
since
Vertices:(, ) and (, )
Co-vertices: (,), (,,)
B2=(, and that passes through foci and }
}
}
}
}
}
}
or
points are:
E1=(, )
E2=(, )
E3=(, )
E4=(, )
6. Asymptotes:
=>=>
=>=>
7. Eccentricity:

≈
8. Length of LR:
The length of the Latus Rectum is
9. Length of Conjugate Axis:
10. Length of Traverse Axis:

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