Need help having trouble with this question....

$\displaystyle x=2cos\theta, y=\frac{1}{\sqrt2}sin\theta$

a) Find gradient at the point where theta=pi

b) Hence find the cartesian co-ordinates

c) Find equations of the tangent and normal at this point.

My workings

a)

$\displaystyle \frac{dx}{d\theta}=-2sin\theta , \frac{dy}{d\theta}=\frac{1}{\sqrt2}cos\theta$

$\displaystyle \frac{d\theta}{dx}=-\frac{1}{2sin\theta}$

$\displaystyle \frac{dy}{dx}=\frac{dy}{d\theta} x \frac{d\theta}{dx}$

$\displaystyle \frac{dy}{dx}=\frac{1}{\sqrt2}cos\theta x -\frac{1}{2sin\theta}$

$\displaystyle \frac{dy}{dx}=-\frac{\frac{1}{\sqrt2}cos\theta}{2sin\theta}$

$\displaystyle \frac{dy}{dx}=-\frac{\frac{1}{\sqrt2}}{2tan\theta}$

$\displaystyle \frac{dy}{dx}=-\frac{\frac{1}{\sqrt2}}{2tan\pi}$

as tan pi = 0, i will get an undefined answer is so how can i do the following parts? :s can someone please help me out

And yes Theta=Pi heres the original question, Question 3 here [/URL]