The equation of the normal to the hyperbola x = 4t and y = 4/t at the point (8, 2) is y = 4x - 30. Find the coordinates of the point where this normal crosses the curve again.
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So 4/t=16t - 30 or 4=16t^2 - 30t or 16t^2 - 30t - 4 = 0 or 8t^2 - 15t - 2 = 0 or (8t + 1)(t - 2)=0
So t =2 (first given solution) or -1/8 so the other point is 4.-1/8, 4/-1/8 or (-1/2,-32).
So t =2 (first given solution) or -1/8 so the other point is 4.-1/8, 4/-1/8 or (-1/2,-32).
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Simply substitute x = 4t and y = 4/t into the line equation y = 4x - 30 and solve the resulting quadratic. One result will be t = 2 corresponding to the point (8, 2). You want the other one.