document.write( "Question 1136852: Hi. I am having a very tough time with the conic sections portion of my algebra class. Is there anyway someone can help me solve the equation:\r
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document.write( "Find the center, transverse axis, vertices, and foci of the hyperbola: y^2/81 - x^2/64 = 1\r
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document.write( "any help would be sincerely grateful!
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document.write( "Thank you in advance for any help! \n" );
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Algebra.Com's Answer #754680 by greenestamps(13200)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "(1) Determine the center. \n" ); document.write( "The standard form of the equation is \n" ); document.write( " \n" ); document.write( "In this example, since the numerators are y^2 and x^2, h and k are both 0; so the center of the hyperbola is the origin, (0,0). \n" ); document.write( "(2) Determine whether the branches open up and down, or right and left. \n" ); document.write( "In my experience, most students (and most teachers!) prefer to use a rule -- something like \"the y^2 term is first, so the branches open in the y direction\". \n" ); document.write( "I prefer to use my UNDERSTANDING of what is going on, rather than a rule which might be easy to forget. Look what happens if you try to choose y=0. You end up with -x^2/64 = 1, which is not possible for real numbers. That means there is no point on the graph where y=0; and that means the branches open up and down. \n" ); document.write( "(3) Transverse axis and vertices \n" ); document.write( "The transverse axis is the line that passes through the points on the two branches of the hyperbola that are closest to each other. With center at the origin and the branches opening up and down, the transverse axis is the y-axis -- i.e., the line x = 0. \n" ); document.write( "To find the vertices -- the points on the transverse axis that are points on the hyperbola -- set x=0 in the equation, giving you y = 9 or -9. The two vertices are (0,9) and (0,-9). \n" ); document.write( "(4) Coordinates of the foci \n" ); document.write( "The distance from the center of the hyperbola to each focus is c, where c^2 = a^2+b^2. In this hyperbola, that is c^2 = 81+64 = 145, so c = sqrt(145). Since the branches of the hyperbola open up and down, the coordinates of the foci are (0,sqrt(145) and (0,-sqrt(145). \n" ); document.write( " |