Answer
tangent $y=-\frac{1}{4}x+\frac{9}{4}$
normal $y=4x-2$
Work Step by Step
Step 1. Given $x^2+2y^2=9$, we have $2xdx+4ydy=0$ or $\frac{dy}{dx}=-\frac{x}{2y}$ which gives the slope of tangent lines to the curve.
Step 2. At point $(1,2)$, the slope of the tangent line is $m=-\frac{1}{2(2)}=-\frac{1}{4}$ and the equation is $y-2=-\frac{1}{4}(x-1)$ or $y=-\frac{1}{4}x+\frac{9}{4}$
Step 3. The slope of the normal line is $n=-1/m=4$ and the equation is $y-2=4(x-1)$ or $y=4x-2$
