Solved

Write the Standard Form of the Equation of the Hyperbola (y+3)29−(x+2)236=1\frac { ( y + 3 ) ^ { 2 } } { 9 } - \frac { ( x + 2 ) ^ { 2 } } { 36 } = 1

Question 63

Multiple Choice

Write the standard form of the equation of the hyperbola. Write the standard form of the equation of the hyperbola. A) \frac { ( y + 3 ) ^ { 2 } } { 9 } - \frac { ( x + 2 ) ^ { 2 } } { 36 } = 1 B) \frac { ( x - 2 ) ^ { 2 } } { 36 } - \frac { ( y - 3 ) ^ { 2 } } { 9 } = 1 C) \frac { ( y + 3 ) ^ { 2 } } { 36 } - \frac { ( x + 2 ) ^ { 2 } } { 9 } = 1 D) \frac { ( y - 3 ) ^ { 2 } } { 36 } - \frac { ( x - 2 ) ^ { 2 } } { 9 } = 1 E) \frac { ( x + 2 ) ^ { 2 } } { 36 } - \frac { ( y + 3 ) ^ { 2 } } { 9 } = 1


A) (y+3) 29−(x+2) 236=1\frac { ( y + 3 ) ^ { 2 } } { 9 } - \frac { ( x + 2 ) ^ { 2 } } { 36 } = 1
B) (x−2) 236−(y−3) 29=1\frac { ( x - 2 ) ^ { 2 } } { 36 } - \frac { ( y - 3 ) ^ { 2 } } { 9 } = 1
C) (y+3) 236−(x+2) 29=1\frac { ( y + 3 ) ^ { 2 } } { 36 } - \frac { ( x + 2 ) ^ { 2 } } { 9 } = 1
D) (y−3) 236−(x−2) 29=1\frac { ( y - 3 ) ^ { 2 } } { 36 } - \frac { ( x - 2 ) ^ { 2 } } { 9 } = 1
E) (x+2) 236−(y+3) 29=1\frac { ( x + 2 ) ^ { 2 } } { 36 } - \frac { ( y + 3 ) ^ { 2 } } { 9 } = 1

Correct Answer:

verifed

Verified

Unlock this answer now
Get Access to more Verified Answers free of charge

Related Questions

Q58: Write the standard form of the

Q59: Identify the vertices and center of

Q60: Identify the vertex and focus of

Q61: Write the standard form of the

Q62: Write the standard form of the

Q64: Graph the function <span class="ql-formula"

Q65: Solve the system by the method

Q66: Write the standard form of the

Q67: Identify the vertices and center of

Q68: Write the standard form of the