Question 189361
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You have:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  4x^2 - 25y^2 - 50y - 125= 0]


And you want to obtain:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{(x - h)^2}{a^2} - \frac{ (y - k)^2}{b^2} = 1]


Where (<i>h</i>,<i>k</i>) is the point about which the hyperbola is centered, <i>a</i> is the semi-major axis, and <i>b</i> is the semi-minor axis.


The process involves completing the square separately for the <i>x</i> and <i>y</i> variables.


First, put the constant term on the right:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  4x^2 - 25y^2 - 50y = 125]


Since there is no 1st degree <i>x</i> term, you can factor directly:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  4(x - 0)^2 - 25y^2 - 50y = 125]


Take the lead coefficient on the *[tex \Large y^2] term and factor it out of the <i>y</i> terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  4(x - 0)^2 - 25 (y^2 + 2y +\ \ \ ) = 125]


Divide the coefficient on the 1st order <i>y</i> 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.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  4(x - 0)^2 - 25 (y^2 + 2y + 1 ) = 125 - 25]


Factor the perfect square inside of the parentheses:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  4(x - 0)^2 - 25 (y + 1)^2 = 100]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{4(x - 0)^2}{100} - \frac{25 (y + 1)^2}{100} = 1]


Reduce the fractions:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{(x - 0)^2}{25} - \frac{ (y + 1)^2}{4} = 1]


Express the denominators as the square root squared:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{(x - 0)^2}{5^2} - \frac{ (y + 1)^2}{2^2} = 1]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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