Question

The curve C has equation x= cos2y​/e^y. Find the equation of the tangent to the curve at the point where y=π​/8 in the form ax+by+c=0

Answer 1

`x=\cfrac{cos(2y)}{e^y}\hspace{5em}x\left( (\pi )/(8) \right)=\cfrac{1}{e^{(\pi )/(8)}√(2)} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{dx}{dy}\implies \stackrel{quotient~rule}{\cfrac{-2sin(2y)e^y-cos(2y)e^y}{(e^y)^2}}\implies \cfrac{-2sin(2y)-cos(2y)}{e^y} \\\\[-0.35em] ~\dotfill\\\\ \left. \cfrac{dx}{dy} \right|_{y=(\pi )/(8)}\implies \cfrac{-2sin((\pi )/(4))-cos((\pi )/(4))}{e^y}\implies \cfrac{-3}{e^{(\pi )/(8)}√(2)}\impliedby m`

now, since we have the value of "x" at π/8 and we know that at that point y = π/8, and we also have the slope at π/8, let's plug all those fellows in the point-slope formula

`\begin{array}ll \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\cfrac{\pi }{8}~~ = ~~\cfrac{-3}{e^{(\pi )/(8)}√(2)}\left(x-\cfrac{1}{e^{(\pi )/(8)}√(2)} \right)`

`y-\cfrac{\pi }{8}~~ = ~~\cfrac{-3}{e^{(\pi )/(8)}√(2)}x+\cfrac{3}{2e^{(\pi )/(4)}} \implies \boxed{\cfrac{3}{e^{(\pi )/(8)}√(2)} {\Large \begin{array}{llll} x \end{array}} + {\Large \begin{array}{llll} y \end{array}}-\cfrac{\pi }{8}-\cfrac{3}{2e^{(\pi )/(4)}}~~ = ~~0}`