text: find an equation for the hyperbola with focus (11,2), and asymptotes 4x-3=18 and 4x+3y=30.?   first, plot all the points on a graph. the focus is (11, 2). now, plot the 2 asymptotes. this can be done by substituting, x=0 to find y then y=0 to find x. the first one is a line crossing (0,-6) and (4.5,0) the second one is a line crossing (0,10) and (7.5,0).  now with these lines and the focus, you will see that the hyperbola opens to the right.  ---  the equation is: {[(x-h)^2]/(a^2)} - {[(y-k)^2]/(b^2)} = 1  with the equations of the asymptotic lines as: y - k = +/- (b/a)(x - h)  ----- so the 2 asymptotes equation will give, y = 4x/3 - 6 and y = -4x/3 + 10  hence, b=4, a=3 solving for h, k  k - 4h/3 = -6 k + 4h/3 = 10 now, 2k = 4, this k = 2.  giving h = 6  --- the equation is  {[(x-6)^2]/(3^2)} - {[(y-2)^2]/(4^2)} = 1