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What is the standard form of the equation of a hyperbola whose general form is 9y 2 – 4x 2 – 18y – 16x 29 = 0

User PAULDAWG
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Final answer:

The standard form of the equation of the given hyperbola is (y – 1)^2/4 – (x + 2)^2/8 = 1.

Step-by-step explanation:

The standard form of the equation of a hyperbola is given by:

(x-h)^2/a^2-(y-k)^2/b^2=1

Where (h,k) is the center of the hyperbola and a and b are the distances from the center to the vertices along the transverse and conjugate axes, respectively.

To find the standard form of the given equation, we need to complete the square for both x and y terms.

Starting with the given equation:
9y^2 – 4x^2 – 18y – 16x + 29 = 0

Rearrange the terms:
9y^2 – 18y – 4x^2 – 16x + 29 = 0

Now group the x terms and the y terms:
(9y^2 – 18y) – (4x^2 + 16x) + 29 = 0

Factor out the common factors:
9(y^2 – 2y) – 4(x^2 + 4x) + 29 = 0

Now complete the square for each term:
(y^2 – 2y + 1) – 1 – 4(x^2 + 4x + 4) + 4 + 29 = 0

Simplify:
(y – 1)^2 – 1 – 4(x + 2)^2 + 33 = 0

Combine like terms:
(y – 1)^2 – 4(x + 2)^2 + 32 = 0

The standard form of the equation is:
(y – 1)^2/4 – (x + 2)^2/8 = 1

User Beroza Paul
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