document.write( "Question 621020: y^2-x^2/15=1\r
\n" ); document.write( "\n" ); document.write( "Want to know the focus,vertices and the asymptote \n" ); document.write( "
Algebra.Com's Answer #390561 by KMST(5328)\"\" \"About 
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For hyperbolas, we like the standard form of the equation, the one that shows a difference of squares equal to 1.
\n" ); document.write( "It may look like
\n" ); document.write( "\"x%5E2%2Fa%5E2-y%5E2%2Fb%5E2=1\" or \"y%5E2%2Fa%5E2-x%5E2%2Fb%5E2=1\" .
\n" ); document.write( "That form of the equation shows you all the numbers you need to know to figure out the he foci, vertices and the asymptotes. (All you need is the a and b numbers).
\n" ); document.write( "\"y%5E2-x%5E2%2F15=1\" is the equation (in standard form) of a hyperbola centered at the origin, so this is an easy problem.
\n" ); document.write( "We know this one is centered at the origin because there is just an \"x%5E2\" and a \"y%5E2\", with nothing added or subtracted before squaring.
\n" ); document.write( "Because of that simplicity, it is easy to see that changing x to -x gives you the same equation, meaning that the graph is symmetrical with respect to the y-axis. The same can be said of changing y to -y, and the symmetry with respect to the x-axis.\r
\n" ); document.write( "\n" ); document.write( "For y=0 we would have a negative number equal to 1 \"-x%5E2%2F15=1\" and that cannot be.
\n" ); document.write( "So, we can see that the graph does not touch the x-axis, where y=0. (In fact the graph does not even want to get close to the x-axis)
\n" ); document.write( "On the other hand, y cannot be zero, but x can be zero.
\n" ); document.write( "When \"x=0\", you see that \"y%5E2=1\" , meaning \"y=1\" or \"y=-1\" , so the graph goes through the points (0,1) and (0,-1).
\n" ); document.write( "For all other points, \"x%5E2%3E0\" so \"x%5E2%2F15%3E0\" and \"y%5E2=x%5E2%2F15%2B1%3E1\" , meaning that all the other points are even farther away from the x-axis, where y=0.
\n" ); document.write( "The closest that the hyperbola comes to the x-axis is the points (0,1) and (0,-1) , which are the vertices.
\n" ); document.write( "As x (and y) grow larger in absolute value, \"x%5E2\" and \"y%5E2\" grow larger, and the graph gets closer to the asymptotes.
\n" ); document.write( "A little algebra transforms the equation into one that gives us the equations of the asymptotes:
\n" ); document.write( "\"y%5E2-x%5E2%2F15=1\" --> \"y%5E2%2B1=x%5E2%2F15\" --> \"%28y%5E2%2B1%29%2Fx%5E2=1%2F15\" --> \"y%5E2%2Fx%5E2%2B1%2Fx%5E2=1%2F15\" --> \"y%5E2%2Fx%5E2=1%2F15%2B1%2Fx%5E2\"
\n" ); document.write( "As \"x%5E2\" grows larger, \"1%2Fx%5E2\" grows smaller, and the graph gets closer to the graph for
\n" ); document.write( "\"y%5E2%2Fx%5E2=1%2F15\" which is the graph for the lines
\n" ); document.write( "\"y%2Fx=sqrt%281%2F15%29\" <--> \"y=sqrt%281%2F15%29x\" and
\n" ); document.write( "\"y%2Fx=-sqrt%281%2F15%29\" <--> \"y=-sqrt%281%2F15%29x\" .
\n" ); document.write( "Those lines are the asymptotes.
\n" ); document.write( "Because teachers do not like to see square roots in denominators, we may have to write them as
\n" ); document.write( "\"y=%28sqrt%2815%29%2F15%29x\" and \"y=-%28sqrt%2815%29%2F15%29x\" .
\n" ); document.write( "THE FOCI:
\n" ); document.write( "There foci are at a distance \"c\" from the center of the hyperbola, and the number \"c\" is related to the numbers \"a%5E2\" and \"b%5E2\" in the standard form of the equation by a formula that can be derived using the Pythagorean theorem. It is
\n" ); document.write( "\"c%5E2=a%5E2%2Bb%5E2\"
\n" ); document.write( "In this case, your \"a%5E2\" and \"b%5E2\" are 1 and 15, so
\n" ); document.write( "\"c%5E2=1%2B15\" --> \"c%5E2=16\" --> \"c=4\".
\n" ); document.write( "The center was (0,0) (the origin).
\n" ); document.write( "The vertices were ((0,-1) and (0,1), on the y-axis.
\n" ); document.write( "The foci are on the same line, but at distance 4 from the center/origin, at
\n" ); document.write( "(0,-4) and (0,4). \n" ); document.write( "
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