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Convert the general form of the hyperbola to its stand 9y^(2)-4x^(2)-16x+54y+29=0

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Answer and Step-by-step explanation:

To convert the general form of the hyperbola to its standard form, we need to complete the square for both the x and y terms. Here’s how it’s done:

First, we’ll group the x and y terms together and move the constant term to the right side of the equation:

9y^2 + 54y - 4x^2 - 16x = -29

Next, we’ll factor out the coefficients of the squared terms:

9(y^2 + 6y) - 4(x^2 + 4x) = -29

Now, we’ll complete the square for both the x and y terms by adding (6/2)^2 = 9 to the y terms and (4/2)^2 = 4 to the x terms. We’ll also add these values to the right side of the equation to keep it balanced:

9(y^2 + 6y + 9) - 4(x^2 + 4x + 4) = -29 + 9(9) - 4(4)

Simplifying, we get:

9(y + 3)^2 - 4(x + 2)^2 = 44

Finally, we’ll divide both sides by 44 to get the standard form of the hyperbola:

(y + 3)^2/4.89 - (x + 2)^2/11 = 1

So, the standard form of the given hyperbola is (y + 3)^2/4.89 - (x + 2)^2/11 = 1.

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