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