Finding centre of ellipse using a tangent line? I need to determine the centre coordinates (a, b) of the ellipse given by the equation:

$$\dfrac{(x-a)^2}{9} + \dfrac{(y-b)^2}{16} = 1$$

A tangent with the equation $y = 1 - x$ passes by the point (0, 1) on the ellipse's circumference.
I'm guessing I have to find the implicit derivative first, but I'm not quite sure how to derive the part of the right. According to my calculator, the implicit differentiation is:
$\frac{-16(x-a)}{9(y-b)}$
But I'd really like to try and do this by hand. I'm just really not sure of the steps I need to take to solve this.
Thanks for any help.
 A: If you are trying to implicitly find the derivative of $$\dfrac{(x-a)^2}{9} + \dfrac{(y-b)^2}{16} = 1$$
I would start with multiplying through by $144$ to get $$16(x-a)^2 + 9(y-b)^2 = 144$$
Then take the derivative, term by term. 
$[16(x-a)^2 ]' = 16[(x-a)^2]' = 16 \cdot 2(x-a)^1 \cdot x' = 32(x -a) \cdot 1$, since $x' = 1$.    
Likewise for $y$, we find that $[9(y-b)^2]' = 18(y-b) \cdot y'$
So the derivative of  $16(x-a)^2 + 9(y-b)^2 = 144$ is $$32(x-a) + 18(y-b) \cdot y' = 0$$
Subtracting $32(x-a)$ from both sides and then dividing by $18(y-b)$ will give the same result as your calculator gave.
A: Hints: Follow, understand and prove the following
Since the point $\;(0,1)\;$ is on the ellipse then
$$\frac{a^2}9+\frac{(1-b)^2}{16}=1$$
Now differentiate implicitly:
$$\frac29(x-a)dx+\frac18(y-b)dy=0\implies \frac{dy}{dx}=-\frac{\frac29(x-a)}{\frac18(y-b)}=-\frac{16}9\frac{x-a}{y-b}$$
But we know that 
$$-1=\left.\frac{dy}{dx}\right|_{x=0}=-\frac{16}9\frac{-a}{1-b}$$
Well, now solve the two variable equations you got above...
