Answer
$$y = - \frac{{15}}{{16}}x + \frac{9}{4}$$
Work Step by Step
$$\eqalign{
& \frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{9} = 1;{\text{ }}\left( {\frac{{20}}{3}, - 4} \right) \cr
& {\text{Find }}\frac{{dy}}{{dx}}{\text{ using the implicit differentiation}} \cr
& \frac{d}{{dx}}\left[ {\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{9}} \right] = \frac{d}{{dx}}\left[ 1 \right] \cr
& \frac{{2x}}{{16}} - \frac{{2y}}{9}\frac{{dy}}{{dx}} = 0 \cr
& \frac{x}{8} - \frac{{2y}}{9}\frac{{dy}}{{dx}} = 0 \cr
& - \frac{{2y}}{9}\frac{{dy}}{{dx}} = - \frac{x}{8} \cr
& \frac{{dy}}{{dx}} = \frac{{9x}}{{16y}} \cr
& {\text{Calculate the slope at the point }}\left( {\frac{{20}}{3}, - 4} \right) \cr
& \frac{{dy}}{{dx}} = \frac{{9\left( {20/3} \right)}}{{16\left( { - 4} \right)}} \cr
& m = - \frac{{15}}{{16}} \cr
& {\text{The equation of the tangent line at the given point is}} \cr
& y - {y_1} = m\left( {x - {x_1}} \right) \cr
& y + 4 = - \frac{{15}}{{16}}\left( {x - \frac{{20}}{3}} \right) \cr
& y + 4 = - \frac{{15}}{{16}}x + \frac{{25}}{4} \cr
& y = - \frac{{15}}{{16}}x + \frac{9}{4} \cr
& \cr
& {\text{Graph}} \cr} $$
