Question 1142079
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The x^2 term is positive, so the branches open right and left.  The general form of the equation for that kind of hyperbola is<br>
{{{(x-h)^2/a^2-(y-k)^2/b^2 = 1}}}<br>
In that form, the center is (h,k); a is the distance (in the x direction) from the center to each vertex -- i.e., from the center to each end of the transverse axis; and b is the distance (in the y direction) from the center to each end of the conjugate axis.<br>
With that much information, you can find the coordinates of the center and the two vertices.<br>
The distance (in the x direction) from the center to each focus is c, where {{{c^2 = a^2+b^2}}}.<br>
Now you can find the coordinates of the two foci.<br>
For the asymptotes, set the equation equal to 0 instead of 1 and solve.<br>
{{{(x-h)^2/a^2-(y-k)^2/b^2 = 0}}}<br>
The expression on the left is a difference of squares, so it is easy to solve.<br>
{{{((x-h)/a-(y-k)/b)((x-h)/a+(y-k)/b) = 0}}}<br>
The equations of the two asymptotes come from solving the equations<br>
{{{(x-h)/a-(y-k)/b = 0}}} and {{{(x-h)/a+(y-k)/b)= 0}}}