document.write( "Question 1150043: Write an equation for the hyperbola that has eccentricity 2, center at (0,0), and vertex at (0,8)
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Algebra.Com's Answer #854697 by KMST(5398)\"\" \"About 
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FACTS ABOUT HYPERBOLAS:
\n" ); document.write( "The equation for a hyperbola centered at (h,k) can be written as
\n" ); document.write( "\"%28x-h%29%5E2%2Fp%5E2-%28y-k%29%5E2%2Fq%5E2=%22+%22%2B-+1\" for some pair (p,q) of positive numbers.
\n" ); document.write( "If a given equation can be written as above, you would have the values of h, k, p, and q and the sign for the \"%22+%22%2B-+1\" part.
\n" ); document.write( "For \"%28x-h%29%5E2%2Fp%5E2-%28y-k%29%5E2%2Fq%5E2=-1\" <--> \"%28y-k%29%5E2%2Fq%5E2-%28x-h%29%5E2%2Fp%5E2=1\" <--> \"%28y-k%29%5E2%2Fq%5E2=%28x-h%29%5E2%2Fp%5E2%2B1\" , you see that it must be \"%28y-k%29%5E2%2Fq%5E2%3E=1\" --> \"abs%28y-k%29%3E=q\" --> \"system%28y-k%3E=q%2C%22or%22%2Cy-k%3C=-q%29\" --> \"system%28y%3E=k%2Bq%2C%22or%22%2Cy%3C=k-q%29\" ,
\n" ); document.write( "so the graph will have an upper branch and a lower branch, like this:
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\n" ); document.write( "On the other hand, for \"%28x-h%29%5E2%2Fp%5E2-%28y-k%29%5E2%2Fq%5E2=1\" <--> \"%28x-h%29%5E2%2Fp%5E2=%28y-k%29%5E2%2Fq%5E2%2B1\" , so it must be \"%28x-h%29%5E2%2Fp%5E2%3E=1\" <--> \"abs%28x-h%29%3E=p\" --> \"system%28x-h%3E=p%2C%22or%22%2Cx-h%3C=-p%29\" --> \"system%28x%3E=h%2Bp%2C%22or%22%2Cy%3C=h-p%29\" ,
\n" ); document.write( "so there is a left branch and a right branch to the graph, like this:
\n" ); document.write( ": .
\n" ); document.write( "In either case, the red and green lines are the asymptotes, with slopes \"q%2Fp\" and \"-q%2Fp\" .
\n" ); document.write( "Textbooks would write the equation of a hyperbola as
\n" ); document.write( "either \"%28y-k%29%5E2%2Fa%5E2-%28x-h%29%5E2%2Fb%5E2=1\" for a hyperbola with upper and lower branches with vertices at \"y=k+%2B-+a\" ,
\n" ); document.write( "or \"%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1\" for a hyperbola with upper and lower branches with vertices at \"x=h+%2B-+a\" .
\n" ); document.write( "When you write the equation you have number values instead of letters,
\n" ); document.write( "but your teacher may expect you to use the same letters they use.
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\n" ); document.write( "COMPARING THOSE FACTS WITH YOUR QUESTION:
\n" ); document.write( "The hyperbola in your question has its center at (0,0), and vertex at (0,8), directly above its center by 8 units.
\n" ); document.write( "It is just like the hyperbola with equation \"%28y-k%29%5E2%2Fq%5E2-%28x-h%29%5E2%2Fp%5E2=1\" or \"%28y-k%29%5E2%2Fa%5E2-%28x-h%29%5E2%2Fb%5E2=1\" shown in the top drawing above.
\n" ); document.write( "The center for the hyperbola in the equation above was (h,k), so for the hyperbola in your question, (h,k) is (0,0), with \"h=k=0\" .
\n" ); document.write( "The distance between the center and the vertex above was labeled as \"q\" in the drawing, so for the hyperbola in your question, \"q=8\" or \"a=8\".
\n" ); document.write( "All you need to write your equation is the value of \"p\" or \"b\".
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\n" ); document.write( "ABOUT ECCENTRICITY:
\n" ); document.write( "So, what do we know about eccentricity?
\n" ); document.write( "It is defined for ellipses and hyperbolas as a ratio of distances from the center, distance to a focus divided by distance to a vertex to the focus.
\n" ); document.write( "and for both curves the drawings have right triangles with sides labeled \"p\" , \"q\", and \"c\" , or \"a\" , \"b\", and \"c\" ,where \"c\" is the hypotenuse and the distance from center to focus.
\n" ); document.write( "According to the Pythagorean theorem, for those triangles \"c%5E2=p%5E2%2Bq%5E2\" or \"c%5E2=a%5E2%2Bb%5E2\"
\n" ); document.write( "In the drawing above, the triangle side labeled c is not the line segment from the center to a focus, but I draw an arc of circle to show that the lengths are the same.
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\n" ); document.write( "FINDING \"c\" AND \"b%5E2\"
\n" ); document.write( "With the equation written as \"y%5E2%2F8%5E2-x%5E2%2Fb%5E2=1\" , \"with+a=8\" ,
\n" ); document.write( "the eccentricity would be \"e=c%2Fa\" , so \"2=c%2F8\" --> \"2%2A8=c\" --> \"highlight%28c=16%29\"
\n" ); document.write( "Substituting the values found got \"a\" and \"c\" ,
\n" ); document.write( "\"c%5E2=a%5E2%2Bb%5E2\" becomes \"16%5E2=8%5E2%2Bb%5E2\" --> \"b%5E2=16%5E2-8%5E2=256-64=highlight%2892%29\" , \"b=sqrt%28192%29=8sqrt%283%29\"and the equation for the hyperbola in the question,
\n" ); document.write( "\"y%5E2%2F8%5E2-x%5E2%2Fb%5E2=1\" turns into \"highlight%28y%5E2%2F64-x%5E2%2F192=1%29\"
\n" ); document.write( "The graph is
\n" ); document.write( " . The asymptotes' slopes are \"red%28a%2Fb=8%2F8sqrt%283%29=1%2Fsqrt%283%29=sqrt%283%29%2F3%29\" and \"green%28-a%2Fb=-1%2Fsqrt%283%29=-sqrt%283%29%2F3%29\" .
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