Answer
The equation of the tangent is $2x+y-4=0$.
Work Step by Step
Step-1: Differentiate the following equation with respect to x:
$$\frac{x^2}{2}+ \frac{y^2}{8}=1$$, we get,
$$\frac{2x}{2}+ \frac{2y}{8} \frac{dy}{dx}=0$$
$$\implies\frac{dy}{dx}=-4\frac{x}{y}$$
Step-2: To find the slope of the tangent at point $(1,2)$, we need to put $x=1$ and $y=2$ in the above equation:
$$\bigg[\frac{dy}{dx}\bigg]_{(1,2)} = -4 \times \frac{1}{2}=-2$$
Step-3: Using the point-slope form of the straight line, one can figure out the equation of the required tangent to the given ellipse:
$$y-2=-2(x-1)\implies 2x+y-4=0$$
