document.write( "Question 717075: how can I solve this use the eccentricity of each hyperbola to find its equation in standard form center(4,1),horizontal transverse axis is 12 and eccentricity 4/3 \n" ); document.write( "
Algebra.Com's Answer #440075 by KMST(5328)\"\" \"About 
You can put this solution on YOUR website!
The equation of a hyperbola with a horizontal transverse axis can be written in the form
\n" ); document.write( "\"%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1\"
\n" ); document.write( "I had to look it up (not good with names), but that is what is called the standard form.
\n" ); document.write( "(h,k) is the center
\n" ); document.write( "Center, vertices, and foci are on the horizontal line \"y=k\"
\n" ); document.write( "For the vertices,
\n" ); document.write( "\"%28x-h%29%5E2%2Fa%5E2=1\" --> \"%28x-h%29%5E2=a%5E2\" --> \"abs%28x-h%29=a\"
\n" ); document.write( "They are at distance \"a\" from the center, on line \"y=k\"
\n" ); document.write( "and \"2a\" is the distance between the vertices.
\n" ); document.write( "The segment (and the distance) between the vertices is called the transverse axis.
\n" ); document.write( "
\n" ); document.write( "So far we know
\n" ); document.write( "\"h=4\", \"k=1\", and \"2a=12\" --> \"a=6\"
\n" ); document.write( "
\n" ); document.write( "The eccentricity \"e\" is defined based of the focal distance \"c\"
\n" ); document.write( "(distance from each focus to the center) as
\n" ); document.write( "\"e=c%2Fa\"
\n" ); document.write( "and it turns out that \"a\", \"b\" and \"c\" are related by \"c%5E2=a%5E2%2Bb%5E2\"
\n" ); document.write( "We know \"e=4%2F3\" so
\n" ); document.write( "\"4%2F3=c%2F6\" --> \"c=6%2A4%2F3\" --> \"c=8\"
\n" ); document.write( "Then, plugging that (along with \"a=6\") into \"c%5E2=a%5E2%2Bb%5E2\" we get
\n" ); document.write( "\"8%5E2=6%5E2%2Bb%5E2\" --> \"64=36%2Bb%5E2\" --> \"b%5E2=64-36\" --> \"b%5E2=28\"
\n" ); document.write( "
\n" ); document.write( "Finally, plugging the values given (or very easily found) for \"a\", \"h\", and \"k\", plus the hard earned value for \"b%5E2\" into the standard form we get
\n" ); document.write( "\"highlight%28%28x-4%29%5E2%2F36-%28y-1%29%5E2%2F28=1%29\"
\n" ); document.write( "
\n" );