= Solution
Set $g(x,y)=y-(x-2)^2$. The <gradient> of $g$ never vanishes, so the <Lagrange multiplier> equations $\nabla f=\lambda\nabla g$ apply at a constrained <stationary point>. They are $4x=-2\lambda(x-2)$, $2y=\lambda$, and $y=(x-2)^2$. Eliminating $\lambda,y$ gives
$$
x+(x-2)^3=0.
$$
With $t=x-2$, this becomes $t^3+t+2=(t+1)(t^2-t+2)=0$. The quadratic factor is $(t-1/2)^2+7/4>0$, so the only real possibility is $t=-1$. Therefore
$$
\boxed{(x,y)=(1,1),\qquad f=3,\qquad\lambda=2.}
$$
The <level curve> through this point is the <ellipse> $2x^2+y^2=3$, meeting the coordinate axes at $(\pm\sqrt{3/2},0)$ and $(0,\pm\sqrt3)$. Its <tangent line> at $(1,1)$ has <slope> $-2x/y=-2$, equal to the <derivative> $2(x-2)$ of the constraint <parabola> there. The two <curves> are tangent.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2007/paper-2-constrained-minimum.png]
{title=The smallest elliptical contour meeting the constraint is tangent at the minimum}
{description=The highlighted contour is $2x^2+y^2=3$ and the constraint is $y=(x-2)^2$. Its four axis intercepts and the unique contact point $(1,1)$ are marked.}
The <parabola> touches this <ellipse> from outside; every other point on the constraint lies on a larger contour. An exact check of this geometric argument is
$$
2x^2+(x-2)^4-3=(x-1)^2\bigl[(x-3)^2+4\bigr]\geq0,
$$
with equality only at $x=1$. Thus the point is the unique constrained <global minimum>.
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