document.write( "Question 422826: Find the Foci of the hyperbola defined by the equation.The equation is (x+5)^2/9 -(y+8)^2/64 = 1.Please help I have no Idea how to do this. \n" ); document.write( "
Algebra.Com's Answer #295716 by lwsshak3(11628)![]() ![]() ![]() You can put this solution on YOUR website! Find the Foci of the hyperbola defined by the equation.The equation is (x+5)^2/9 -(y+8)^2/64 = 1.Please help I have no Idea how to do this.\r \n" ); document.write( "\n" ); document.write( ".. \n" ); document.write( "(x+5)^2/9-(y+8)^2/64=1 \n" ); document.write( "Standard form of a hyperbola: \n" ); document.write( "y=(x-h)^2/a^2-(y-k)^2/b^2=1, with (h,k) being the (x,y) coordinates of the \n" ); document.write( "center. If the x^2 term is listed first, the hyperbola would have a horizontal transverse axis. If the y^2 term is listed first, the hyperbola would have a vertical transverse axis. \n" ); document.write( ".. \n" ); document.write( "Given hyperbola has a center at (-5,-8) and a horizontal transverse axis, that is, the hyperbola opens sideways. \n" ); document.write( "a^2=9 \n" ); document.write( "a=3 \n" ); document.write( "b^2=64 \n" ); document.write( "b=8 \n" ); document.write( "c^2=a^2+b^2 \n" ); document.write( "c=sqrt(9+64)=sqrt(73)=8.54 \n" ); document.write( "The foci is on the transverse axis, y=-8, \n" ); document.write( "Its coordinates are:(-5+-c,-8) or (13.54,-8) and (3.54,-8) \n" ); document.write( "The graph below might help you understand the problem better. \n" ); document.write( ".. \n" ); document.write( "y=-8+-(64((x+5)^2/9-1))^.5 \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |