If the tangent at the point (2 , 3 ) to the hyperbola x^2 4 - y^2 9 =1 is parallel to 3 x-y+4=0 , then the value of is

Step-by-step solution

The equation x^2 4 - y^2 9 =1 corresponds to the standard hyperbola form with a^2=4 and b^2=9 . The parametric point (2 , 3 ) lies on this hyperbola. The tangent to the hyperbola at this point is given by x 2 4 - y 3 9 = 1 , which simplifies to x 2 - y 3 = 1 . Rearranging to slope-intercept form yields y = 3 2 x - 3 , so the slope is m_t = 3 2 . Since the tangent is parallel to the line 3x - y + 4 = 0 with slope 3, we set 3 2 = 3 , leading to = 1 2 . The value of is 30^ .