Finding centre of ellipse using a tangent line?

Question

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.

Answer 1

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...

Answer 2

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.