Answer
$y=-x+2 \sqrt 2$ and $-\sqrt 2$
Work Step by Step
Here,we have $x'=-2 \sin t; y'=2 \cos t$
and $\dfrac{dy}{dx}=-\cot t $
Also, we have $x=\sqrt 2$ and $y=\sqrt 2 $
$\dfrac{dy}{dx}(t=\dfrac{\pi}{4})=-\cot \dfrac{\pi}{4}=-1$
Thus, $y-\sqrt 2=(-1) (x-\sqrt 2)$
or, $y=-x+2 \sqrt 2$
Now, $\dfrac{d^2y}{dx^2}=\dfrac{\dfrac{dy'}{dt}}{\dfrac{dx}{dt}}=(-\dfrac{1}{2})(\csc^3 t)$
$\dfrac{d^2y}{dx^2}(t=\dfrac{\pi}{4})=(-\dfrac{1}{2})\csc^3 (\dfrac{\pi}{4}) \implies \dfrac{d^2y}{dx^2}(t=\dfrac{\pi}{4})=\dfrac{-1}{2\sin^3 (\dfrac{\pi}{4})}=\dfrac{-1}{2(\dfrac{\sqrt 2}{2})^3}$
Hence, $\dfrac{d^2y}{dx^2}=\dfrac{-1}{2(\dfrac{\sqrt 2}{2})^3}$=-\sqrt 2$
