The normal to the curve x*(e^-y) + e^y = 1 + x, at the point (c,lnc), has a y-intercept c^2 + 1. Determine the value of c.

We firstly need to find the slope of the line normal to the curve at the point (c, lnc). To do this, we will differentiate implicitly the equation of the curve to find the slope of the tangent. By implicit differentiation we get : (e^-y )- x*(e^-y)dy/dx + (e^y)dy/dx = 1. By rearranging we get : slope of tangent = dy/dx = (1-(e^-y))/(e^y -x(e^-y)) (1). Since we want to find the slope of the tangent at the point (c,lnc) we plug into (1) x = c and y = lnc. We get that dy/dx = 1/c. We know that slope of normal = -1/slope of tangent, so we get that slope of normal at (c,lnc) = -c.We will now use the value of the y-intercept given. The equation of a line passing through a point, say (a,b) is given by y-b = slope(x-b). We have that the normal line at (c,lnc) has slope -c and also passes through the point (0, c^2 + 1). So we get that lnc - c^2 - 1 = (-c)* (c) and since two terms cancel out, we have that lnc = 1 implying that c = e.