#### Answer

The tangent is $$y=-\frac{1}{2}x+1$$ and the normal is $$y=2x+1$$

#### Work Step by Step

$$e^x+y^2=2\hspace{2cm}A(0,1)$$
a) Find the derivative $y'$ with respect to $x$, using implicit differentiation: $$e^x+2yy'=0$$ $$y'=-\frac{e^x}{2y}$$
b) The slope of the tangent line at $A(0,1)$ is $$y'=-\frac{e^0}{2\times1}=-\frac{1}{2}$$
The tangent to the curve at $(0,1)$ is $$y-1=-\frac{1}{2}(x-0)$$ $$y-1=-\frac{1}{2}x$$ $$y=-\frac{1}{2}x+1$$
c) The normal is the perpendicular line to the tangent at a point, meaning the product of their slopes is $-1$.
So if we call the slope of the normal at $A(0,1)$ $s$, we have $$s\Big(-\frac{1}{2}\Big)=-1$$ $$s=2$$
The normal to the curve at $(0,1)$ is $$y-1=2(x-0)$$ $$y-1=2x$$ $$y=2x+1$$