SOLUTION: Find the standard form of the hyperbola given by the equation 4x^2 - 25y^2 - 50y - 125= 0 Please help! Thank you!

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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) About Me  (Show Source):
You can put this solution on YOUR website!


You have:



And you want to obtain:



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:



Take the lead coefficient on the term and factor it out of the y terms:



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.



Factor the perfect square inside of the parentheses:



Divide by the constant term on the right so that the right hand side becomes 1:



Reduce the fractions:



Express the denominators as the square root squared:



John