Question 1185475
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With a horizontal transverse axis, the general equation is<br>
{{{(x-h)^2/a^2-(y-k)^2/b^2=1}}}<br>
(h,k) is the center of the hyperbola
2a is the length of the transverse axis (i.e., a is the distance from the center to each vertex)
2b is the length of the conjugate axis
The slopes of the asymptotes are b/a and -b/a<br>
To find the center (h,k), note that the two asymptotes intersect at the center of the hyperbola.  Solve the given pair of equations of the asymptotes to find h and k.<br>
You should find k=3, which means the given point (0,3) is one of the vertices of the hyperbola.  So the distance from the center (h,k) to the vertex (0,3) is a -- half of the transverse axis.<br>
Now you know a, so you know a^2; knowing the slopes 1 and -1 are b/a and -b/a, you can determine b^2.<br>
And now you have all the pieces you need to write the equation.<br>
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ANSWER (but if you want to learn anything from this you should go through the steps described above to find it):<br>
{{{(x-2)^2/4-(y-3)^2/4=1}}}<br>
A graph, showing the two branches of the hyperbola and the two asymptotes....<br>
{{{graph(400,400,-4,8,-4,8,sqrt((x-2)^2-4)+3,-sqrt((x-2)^2-4)+3,x+1,5-x)}}}<br>