Find the coordinates of the stationary point of the curve with equation $(x+y-2)^2 = e^y - 1$ - AQA - A-Level Maths Pure - Question 6 - 2018 - Paper 2

Rearranging the differentiation leads to:

$2(x+y-2) + 2(x+y-2)\frac{dy}{dx} = e^y \frac{dy}{dx}$

For stationary points, we set the derivative rac{dy}{dx} = 0:

$2(x+y-2) = 0$

This simplifies to:

$x + y - 2 = 0 \Rightarrow y = 2 - x$

Now, substituting $y = 2 - x$ back into the original curve's equation:

$(x + (2 - x) - 2)^2 = e^{(2 - x)} - 1$

This simplifies to:

$0 = e^{(2 - x)} - 1 \Rightarrow e^{(2 - x)} = 1$

Taking the natural logarithm, we find:

$2 - x = 0 \Rightarrow x = 2$

Substituting $x = 2$ back into $y = 2 - x$, gives:

$y = 2 - 2 = 0$

Thus, the coordinates of the stationary point are:

Coordinates: $(2, 0)$