document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=428893>Question 428893</A>: <I><SPAN STYLE=\"word-break: break-all;\"> find the vertices,foci and equation of asymptotes of x^2-2y^2=2 for the hyperbola </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #298612 by </B><A HREF=/tutors/aboutme.mpl?userid=lwsshak3><B>lwsshak3(11628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=lwsshak3><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=298612\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Find the vertices,foci and equation of asymptotes of x^2-2y^2=2 for the hyperbola<BR>\n" );
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document.write( "Standard form of hyperbola with horizontal transverse axis(opens sideways):<BR>\n" );
document.write( "(x-h)^2a^2-(y-k)^2/b^2=1<BR>\n" );
document.write( "Standard form of hyperbola with Vertical transverse axis(opens up and down):<BR>\n" );
document.write( "(y-k)^2a^2-(x-h)^2/b^2=1<BR>\n" );
document.write( "(h,k)=(xy) coordinates of the center for both forms<BR>\n" );
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document.write( "Given:<BR>\n" );
document.write( "x^2-2y^2=2<BR>\n" );
document.write( "=x^2/2-y^2/1=1<BR>\n" );
document.write( "This is a hyperbola with a horizontal transverse axis and center at (0.0)<BR>\n" );
document.write( "a^2=2<BR>\n" );
document.write( "a=sqrt(2)<BR>\n" );
document.write( "b^2=1<BR>\n" );
document.write( "b=1<BR>\n" );
document.write( "c^2=a^2+b^2=2+1=3<BR>\n" );
document.write( "c=sqrt(3)<BR>\n" );
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document.write( "The vertices are on the transverse axis located +- a from the center or (0,sqrt(2)) & (0,-sqrt(2))<BR>\n" );
document.write( "Foci are also on the transverse axis just inside the vertices locate =-c from the center or (0,sqrt(3)) & (0,-sqrt(3))<BR>\n" );
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document.write( "When figuring the asymptotes you can think of  straight lines with slope+- b/a for hyperbolas with horizontal transverse axis and slope +-b/a for hyperbolas with vertical transverse axis. And they all go thru the center of the hyperbola. For this reason, you can use the standard form of a straight line, y=mx+b, to get the equations of the hyperbolas. knowing the slope and and a point, the center, on the line. In the given case, equations of the asymptotes are: y=sqrt(2)/2x and y=-sqrt(2)/2x (note that in this case the y-intercept, b=0)<BR>\n" );
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document.write( "ans:<BR>\n" );
document.write( "vertices:(0,sqrt(2)) & (0,-sqrt(2))<BR>\n" );
document.write( "Foci: (0,sqrt(3)) & (0,-sqrt(3))<BR>\n" );
document.write( "equation of asymptotes: y=.707x and y=-.707x<BR>\n" );
document.write( "The graph below can serve as a check on these answers:<BR>\n" );
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document.write( "y=+-((x^2-2)/2)^.5<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+graph%28+300%2C+300%2C+-5%2C+5%2C+-5%2C+5%2C+%28%28x%5E2-2%29%2F2%29%5E.5%2C-%28%28x%5E2-2%29%2F2%29%5E.5%2C.707x%2C-.707x%29+ ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" >  </SPAN>\n" );
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