document.write( "Question 628918: A focal point of the elipse ((x+3)^2)/16)+(y-1)^2)/4)=1 is:
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document.write( "a.) (-1,1)
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document.write( "b.) (-3-2 radical 3, 1)
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document.write( "c.) (-7,1)
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document.write( "d.) 3+2 radical 3, 1)\r
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document.write( "* note: when ever i have entered the equation in to graph it it keeps saying 1/16 (x+3) but that is not correct it is x+3^2 all over 16. \r
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document.write( "I have worked it some way but I dont know how to continue it. I have:\r
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document.write( "Centre: (-3,1)
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document.write( "a=16
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document.write( "b=3
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document.write( "c=15.716\r
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document.write( "but i dont know how to find the focal point. \n" );
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Algebra.Com's Answer #395895 by lwsshak3(11628)![]() ![]() ![]() You can put this solution on YOUR website! Find a focal point of the ellipse: \n" ); document.write( "((x+3)^2)/16)+(y-1)^2)/4)=1 \n" ); document.write( "This is an equation of an ellipse with horizontal major axis: \n" ); document.write( "Its standard form: (x-h)^2/a^2+(y-k)^2/b^2=1, (a>b), (h,k)=(x,y) coordinates of center \n" ); document.write( "For given ellipse: \n" ); document.write( "center(-3,1) \n" ); document.write( "a^2=16 \n" ); document.write( "a=√16=4 \n" ); document.write( "b^2=4 \n" ); document.write( "b=√4=2 \n" ); document.write( "c^2=a^2-b^2=16-12=4 \n" ); document.write( "c=√4=2 \n" ); document.write( "Foci: (Both focal points)=(-3±c,1)=(-3±2,1)=(-5,1) and (-1,1) \n" ); document.write( "answer a is the correct answer \n" ); document.write( " |