Answer
Tangent line: $\quad y=-\displaystyle \frac{1}{2}x+2\sqrt{2}$
$\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=\frac{\pi}{4}}=-\frac{\sqrt{2}}{4}$
Work Step by Step
$\displaystyle \left\{\begin{array}{l}
\frac{dx}{dt}=4\cos t,\\
\\
\frac{dy}{dt}=-2\sin t
\end{array}\right.\quad, \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{-2\sin t}{4\cos t}=-\frac{1}{2}\tan t$
The slope of the tangent line at $t=\displaystyle \frac{\pi}{4}$ is
$\left. \displaystyle \frac{dy}{dx} \right|_{t=\frac{\pi}{4}}= \left. -\displaystyle \frac{1}{2}\tan t \right|_{t=\frac{\pi}{4}}=-\frac{1}{2}\tan\frac{\pi}{4}=-\dfrac{1}{2}$
With $y'=\displaystyle \frac{dy}{dx}$, we find $\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=\frac{\pi}{4}}$,
$\displaystyle \frac{d^{2}y}{dx^{2}}=\frac{dy'/dt}{dx/dt}=\frac{\frac{d}{dt}[-\frac{1}{2}\tan t]}{4\cos t}$
$=\displaystyle \frac{\frac{1}{2}\sec^{2}t}{4\cos t}=-\frac{1}{8\cos^{3}t}$
$\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=\frac{\pi}{4}}=-\frac{1}{8\cos^{3}\frac{\pi}{4}}=-\frac{1}{8(\frac{1}{\sqrt{2}})^{3}}=-\frac{2\sqrt{2}}{8}=-\frac{\sqrt{2}}{4}$
The point $P$ on the curve corresponding to $t=\displaystyle \frac{\pi}{4}$ is
$\left\{\begin{array}{l}
x=4\sin\frac{\pi}{4}=2\sqrt{2}\\
\\
y=2\cos\frac{x}{4}=\sqrt{2}
\end{array}\right.\quad, P(2\sqrt{2},\sqrt{2})$
Use the point-slope equation for the tangent line.
$y-\displaystyle \sqrt{2}=-\frac{1}{2}(x-2\sqrt{2})$
$y=-\displaystyle \frac{1}{2}x+\sqrt{2} +\sqrt{2}$
$y=-\displaystyle \frac{1}{2}x+2\sqrt{2}$
