document.write( "Question 834063: Find the vertices and foci of the hyperbola.
\n" ); document.write( "16x2 − y2 − 64x − 4y + 44 = 0
\n" ); document.write( "

Algebra.Com's Answer #503045 by lwsshak3(11628)\"\" \"About 
You can put this solution on YOUR website!
Find the vertices and foci of the hyperbola.
\n" ); document.write( "16x2 − y2 − 64x − 4y + 44 = 0
\n" ); document.write( "***
\n" ); document.write( "16x^2-64x-y^2+4y=-44
\n" ); document.write( "hyperbola has a horizontal transverse axis.
\n" ); document.write( "Its standard form of equation:\"%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1\"
\n" ); document.write( "complete the square
\n" ); document.write( "16(x^2-4x+4)-(y^2-4y+4)=-44+64-4
\n" ); document.write( "16(x-2)^2-(y-2)^2=16
\n" ); document.write( "equation:\"%28x-2%29%5E2-%28y-2%29%5E2%2F16=1\"
\n" ); document.write( "center: (2,2)
\n" ); document.write( "a^2=1
\n" ); document.write( "a=1
\n" ); document.write( "vertices: (2±a,2)=(2±1,2)=(1,2) and (3,2)
\n" ); document.write( "b^2=16
\n" ); document.write( "b=4
\n" ); document.write( "c^2=a^2+b^2=1+16=17
\n" ); document.write( "c=√17≈4.1
\n" ); document.write( "foci:(2±c,2)=(2±4.1,2)=(-2.1,2) and (6.1,2)\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "
\n" ); document.write( "
\n" );