If $m_{1}and{m}_2$ the slopes of the tangent to the hyperbola $\frac{x^2}{25}-\frac{y^2}{16}=1$ which pass through the point (6, 2) Then

The line through (6,2) is

$y-2=m\left(x-6\right)\Rightarrow y=mx+2-6m$

Now from condition of tangency, ${\left(2-6m\right)}^2=25m^2-16$

$\Rightarrow 36m^2+4-24m-25m^2+16=0$

$\Rightarrow 11m^2-24m+20=0$

 Obviously its roots are $m_1$ and $m_2$ , therefore

 $m_1+m_2=\frac{24}{11}and{m}_1{m}_2=\frac{20}{11}$