#### Answer

The tangent is $$y=-\frac{1}{4}x+\frac{17}{4}$$ and the normal is $$y=4x$$

#### Work Step by Step

$$x^{3/2}+2y^{3/2}=17\hspace{2cm}(1,4)$$
a) Find the derivative of the function: $$\frac{3}{2}x^{1/2}+2\times\frac{3}{2}y^{1/2}y'=0$$ $$x^{1/2}+2y^{1/2}y'=0$$ $$y'=-\frac{x^{1/2}}{2y^{1/2}}=-\frac{1}{2}\sqrt{\frac{x}{y}}$$
b) The slope of the tangent at $(1,4)$ is $$y'=-\frac{1}{2}\sqrt{\frac{1}{4}}=-\frac{1}{2}\times\frac{1}{2}=-\frac{1}{4}$$
The tangent to the curve at $(1,4)$ is $$y-4=-\frac{1}{4}(x-1)$$ $$y-4=-\frac{1}{4}x+\frac{1}{4}$$ $$y=-\frac{1}{4}x+\frac{17}{4}$$
c) Since the normal is the perpendicular line to the tangent at a point, the product of their slopes equals $-1$.
Therefore, if we call the slope of the normal at $(1,4)$ $s$, we would have $$s\times\Big(-\frac{1}{4}\Big)=-1$$ $$s=4$$
The normal to the curve at $(1,4)$ is $$y-4=4(x-1)$$ $$y-4=4x-4$$ $$y=4x$$