document.write( "Question 572346: Please help me solve and graph this equation: $\"+9x%5E2-4y%5E2%2B18x%2B32y-91=0+\"$ \r
\n" ); document.write( "\n" ); document.write( "I tried putting it in standard form and it ended up being a horizontal hyperbola. The standard form was $\"+%28x%2B2%29%5E2+%2F+7+-+4%28y%2B4%29%5E2+%2F+63+=+1+\"$ \r
\n" ); document.write( "\n" ); document.write( "From there I got the center of the hyperbola which is (h,k) or (-2,-4)and I graphed it. Then I looked for the values for a, b, and c. A was $\"+sqrt+%28+7+%29+\"$ or 2.6 as a decimal and B was $\"+sqrt+%28+63+%29+\"$ or 7.9 as a decimal. I graphed A and B as the conjugate and transverse axes. This is where I got confused. The equation for the hyperbola had x on the left and y on the right, meaning it should graph horizontally. But I graphed the axes and the hyperbola came out vertical. Please help me see where I went wrong! \n" ); document.write( "

Algebra.Com's Answer #368436 by KMST(5328) $\"\"$ $\"About$

You can put this solution on YOUR website!
Here's how I go about it:
\n" ); document.write( " $\"+9x%5E2-4y%5E2%2B18x%2B32y-91=0+\"$ --> $\"%289x%5E2%2B18x%29%2B%28-4y%5E2%2B32y%29-91=0\"$ --> $\"9%28x%5E2%2B2x%29%2B%28-4%29%28y%5E2-8y%29-91=0\"$ --> $\"9%28%28x%5E2%2B2x%2B1%29-1%29-4%28%28y%5E2-8y%2B16%29-16%29-91=0\"$ --> $\"9%28%28x%2B1%29%5E2-1%29-4%28%28y-4%29%5E2-16%29-91=0\"$ --> $\"9%28x%2B1%29%5E2-9-4%28y-4%29%5E2-4%28-16%29-91=0\"$ --> $\"9%28x%2B1%29%5E2-4%28y-4%29%5E2-9%2B64-91=0\"$ --> $\"9%28x%2B1%29%5E2-4%28y-4%29%5E2-36=0\"$ --> $\"9%28x%2B1%29%5E2-4%28y-4%29%5E2=36\"$
\n" ); document.write( "and dividing both sides by 36
\n" ); document.write( " $\"%28x%2B1%29%5E2%2F4-%28y-4%29%5E2%2F9=1\"$ \n" ); document.write( "

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