Answer
tangent $y=-\frac{1}{4}x+\frac{17}{4}$
normal $y=4x$
Work Step by Step
Step 1. Given $x^{3/2}+2y^{3/2}=17$, we have $\frac{3}{2}x^{1/2}+3y^{1/2}y'=0$ or $y'=-\frac{\sqrt x}{2\sqrt y}$ which gives the slope of tangent lines to the curve.
Step 2. At point $(1,4)$, the slope of the tangent line is $m=-\frac{\sqrt 1}{2\sqrt 4}=-\frac{1}{4}$ and the equation is $y-4=-\frac{1}{4}(x-1)$ or $y=-\frac{1}{4}x+\frac{17}{4}$
Step 3. The slope of the normal line is $n=-1/m=4$ and the equation is $y-4=4(x-1)$ or $y=4x$
