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

Using implicit differentiation, we can isolate $\frac{dy}{dx}$ in the equation:

$2(x+y-2)\left(1+\frac{dy}{dx}\right)=e^{y}\frac{dy}{dx}$

Let's rearrange this to bring all terms involving $\frac{dy}{dx}$ to one side. This leads to:

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

We can factor out $\frac{dy}{dx}$:

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

Setting $\frac{dy}{dx}=0$ gives:

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

Thus:

$x+y-2=0$

This implies:

$y=2-x$

Substituting $y=2-x$ back into the original equation:

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

This simplifies to:

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

Thus:

$e^{2-x}=1$

Taking the natural logarithm gives:

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

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

$y=2-2=0$

Thus, the coordinates of the stationary point are (2, 0).

The value of $x$ has been determined as:

$x=2$

We achieve the final result by concluding that the coordinates of the stationary point are:

$(x, y) = (2, 0)$