document.write( "Question 309219: Write an equation for the hyperbola...I have no idea.
\n" ); document.write( "vertices (0,6) and (0,-6) conjugate axis of 14.
\n" ); document.write( "I'd really like to figure out on my own but I have no idea where to start..so if someone could give me a step-by-step solution, that would be wonderful! Thank you. \n" ); document.write( "

Algebra.Com's Answer #221108 by jim_thompson5910(35256) $\"\"$ $\"About$

You can put this solution on YOUR website!
The center of the hyperbola is the midpoint of the line segment from vertex to vertex. So the midpoint of (0,6) and (0,-6) is ((0+0)/2, (6+(-6)/2) ---> (0, 0). So the center is (0,0). The center is in the form (h,k), so h=0 and k=0.\r
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\n" ); document.write( "\n" ); document.write( "In this case, 'a' is the length of the semi-minor axis and 'b' is the length of the semi-major axis. So the lengths of the conjugate and traverse axes are 2a and 2b units respectively.\r
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\n" ); document.write( "\n" ); document.write( "Since the traverse axis is 2b units long, this means that 2b=12 (since the distance between the vertices is 12 units) which means that b=6. Also, because the conjugate axis is 14 units long, and 2a is the length of the conjugate axis, this means that 2a=14 and a=7\r
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\n" ); document.write( "\n" ); document.write( "Finally, recall that the general equation for a hyperbola which opens up vertically is $\"%28%28y-k%29%5E2%29%2F%28b%5E2%29-%28%28x-h%29%5E2%29%2F%28a%5E2%29=1\"$ . From here, all you need to do is plug in the right values and simplify. \n" ); document.write( "

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