Answer
Equation of tangent:
$$3x-2y-10=0$$
Equation of Normal:
$$ 2x+3y-24=0$$
Work Step by Step
Step-1: We need to find slope of the curve at $(6,4)$. So, we will differentiate $x^2 - y^2 = 20$ with respect to $x$,
$$2x - 2y\frac{dy}{dx} = 0$$
$$\implies \frac{dy}{dx} = \frac{x}{y}$$
At $(6, 4)$, slope $m = \frac{dy}{dx} = 3/2$.
Step-2: Using point-slope form, we determine the equation of the tangent,
$$y-4 = \frac{3}{2}(x-6)\implies3x-2y-10=0$$
Step-3: Let the slope of the normal be $m_n$. Thus, $m_n \times m = -1\implies m_n = -2/3$. Using the slope-point form, we will determine the equation of normal.
$$y-4=-\frac{2}{3}(x-6)\implies 2x+3y-24=0$$
Step-4: Picture of the graph is given below.