document.write( "Question 706005: how can i find the focus/foci and vertex/vertices\r
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document.write( "x^2-4y^2-2x+16y=20\r
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document.write( "this is how far i have gotten, but i am not sure if it is right so far\r
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document.write( "x^2-4y^2-2x+16y=20
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document.write( "(x^2-2x+1)-4(y^2+4y+4)=20+1+4
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document.write( "(x-1)^2/25-4(y+4)^2/25=25/25
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document.write( "(x-1)^2/25-4(y+4)^2/25=1\r
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document.write( "please help. thank you. \n" );
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Algebra.Com's Answer #434927 by lwsshak3(11628)![]() ![]() ![]() You can put this solution on YOUR website! how can i find the focus/foci and vertex/vertices \n" ); document.write( "x^2-4y^2-2x+16y=20 \n" ); document.write( "complete the square: \n" ); document.write( "x^2-2x-4y^2+16y=20 \n" ); document.write( "(x^2-2x+1)-4(y^2-4y+4)=20+1-16 \n" ); document.write( "(x-1)^2-4(y-2)^2=5 \n" ); document.write( " \n" ); document.write( "This is an equation of a hyperbola with horizontal transverse axis. \n" ); document.write( "Its standard form: \n" ); document.write( "For given equation: \n" ); document.write( "Center: (1,2) \n" ); document.write( "a=5 \n" ); document.write( "a^2=25 \n" ); document.write( "vertices: (1±a,2)=(1±5,2)=(-4,2) and (6,2) \n" ); document.write( ".. \n" ); document.write( "b=5/4 \n" ); document.write( "b^2=25/16 \n" ); document.write( ".. \n" ); document.write( "foci \n" ); document.write( "c^2=a^2+b^2=25+(25/16)=425/16 \n" ); document.write( "c=√425/4≈5.15 \n" ); document.write( "foci:(1±c,2)=(1±5.15,2)=(-4.15,2) and (6.15,2)\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |