SOLUTION: Find the standard form of the hyperbola given by the equation 4x^2 - 25y^2 - 50y - 125= 0
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Question 189361: Find the standard form of the hyperbola given by the equation 4x^2 - 25y^2 - 50y - 125= 0
Please help!
Thank you!
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
You have:
And you want to obtain:
^2}{a^2} - \frac{ (y - k)^2}{b^2} = 1)
Where (h,k) is the point about which the hyperbola is centered, a is the semi-major axis, and b is the semi-minor axis.
The process involves completing the square separately for the x and y variables.
First, put the constant term on the right:
Since there is no 1st degree x term, you can factor directly:
^2 - 25y^2 - 50y = 125)
Take the lead coefficient on the 
term and factor it out of the y terms:
^2 - 25 (y^2 + 2y +\ \ \ ) = 125)
Divide the coefficient on the 1st order y term, square the result, and add to both sides. Remember, since you have the factor of -25 outside of the parentheses, you are actually adding -25 to both sides.
^2 - 25 (y^2 + 2y + 1 ) = 125 - 25)
Factor the perfect square inside of the parentheses:
^2 - 25 (y + 1)^2 = 100)
Divide by the constant term on the right so that the right hand side becomes 1:
^2}{100} - \frac{25 (y + 1)^2}{100} = 1)
Reduce the fractions:
^2}{25} - \frac{ (y + 1)^2}{4} = 1)
Express the denominators as the square root squared:
^2}{5^2} - \frac{ (y + 1)^2}{2^2} = 1)
John
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