Answer
(a) Equation of normal at $(4, 2) \rightarrow 2x-y-6=0$.
(b) See the given picture.
(c) $(1.647, -2.706)$ or $\big(\frac{28}{17}, -\frac{46}{17}\big) \text{ and } (4,2))$
Work Step by Step
Step-1: Differentiate the following equation with respect to $x$:
$$\frac{x^2}{32}+ \frac{y^2}{8}=1$$,
we get,
$$\frac{2x}{32} + \frac{2y}{8} \frac{dy}{dx} = 0$$
$$\implies \frac{dy}{dx}=-\frac{x}{4y}$$
Step-2: Find the slope of the ellipse at $(4,2)$:
$$\bigg[\frac{dy}{dx}\bigg]_{(4,2)}=-\frac{4}{4\times 2}=-\frac{1}{2}$$
Step-3: Let the slope of normal be $m$, therefore:
$$m \times -\frac{1}{2}=-1\implies m=2$$
Step-4: Using the point-slope form, equation of normal can be easily determined:
$$y-2=2(x-4)\implies 2x-y-6=0$$
Step-5: Graph the ellipse and the normal equation on the graphing calculator. It should look something like the picture given below.
Step-6: From the graph, it can be seen that the normal line intersect the ellipse at $(1.647, -2.706)$ or $\big(\frac{28}{17}, -\frac{46}{17}\big) \text{ and } (4,2)$
