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Section 4.4 : Hyperbolas

5. Complete the square on the \(x\) and \(y\) portions of the equation and write the equation into the standard form of the equation of the hyperbola.

\[25{y^2} + 250y - 16{x^2} - 32x + 209 = 0\]

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Start Solution

The process here will be is identical to the process we used in the previous section to write equations of ellipses in standard form.

The first step is to make sure the coefficient of the \({x^2}\) and \({y^2}\) is a one. The \({x^2}\) has a coefficient of -16 and the \({y^2}\) has a coefficient of 25. What we will do is factor a -16 out of every term involving an \(x\) and a 25 out of ever term involving a \(y\). Doing that gives,

\[25\left( {{y^2} + 10y} \right) - 16\left( {{x^2} + 2x} \right) + 209 = 0\] Show Step 2

Now let’s get started on completing the square. First, we need one-half the coefficient of the \(x\) and \(y\) term, square each and the add/subtract those numbers in the appropriate places as follows,

\[\require{color}{\left( {\frac{{10}}{2}} \right)^2} = {\left( 5 \right)^2} = \,{\color{Red} 25}\hspace{0.25in}{\left( {\frac{2}{2}} \right)^2} = {\left( 1 \right)^2} = \,{\color{ProcessBlue} 1}\] \[\require{color}25\left( {{y^2} + 10y \,{\color{Red} + 25 - 25}} \right) - 16\left( {{x^2} + 2x \,{\color{ProcessBlue}+ 1 - 1}} \right) + 209 = 0\]

Be careful you add/subtract these numbers and make sure you put them in the parenthesis!

Show Step 3

Next, we need to factor the \(x\) and \(y\) terms and add up all the constants.

\[\require{color}\begin{align*}25\left( {{{\left( {y + 5} \right)}^2} \,{\color{Red} - 25}} \right) - 16\left( {{{\left( {x + 1} \right)}^2} \,{\color{ProcessBlue}- 1}} \right) + 209 & = 0\\ 25{\left( {y + 5} \right)^2} - 16{\left( {x + 1} \right)^2} + 209 \,{\color{Red} - 625} \,{\color{ProcessBlue}+ 16} & = 0\\ 25{\left( {y + 5} \right)^2} - 16{\left( {x + 1} \right)^2} - 400 & = 0\end{align*}\]

When adding the constants up, make sure to multiply the -16 through the \(x\) terms and the 25 through the \(y\) terms before adding the constants up.

Show Step 4

To finish things off we’ll first move the 400 to the other side of the equation.

\[25{\left( {y + 5} \right)^2} - 16{\left( {x + 1} \right)^2} = 400\]

To get this into standard form we need a one on the right side of the equation. To get this all we need to do is divide everything by 400 and we’ll do a little simplification work on the terms.

\[\frac{{25{{\left( {y + 5} \right)}^2}}}{{400}} - \frac{{16{{\left( {x + 1} \right)}^2}}}{{400}} = 1\hspace{0.25in} \Rightarrow \hspace{0.25in}\require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{{{\left( {y + 5} \right)}^2}}}{{16}} - \frac{{{{\left( {x + 1} \right)}^2}}}{{25}} = 1}}\]