Find a point on the curve $(x − 1)^3 = y^2$ with the minimum distance from the origin

Question

I am asked to find a point on the curve $(x − 1)^3 = y^2$ with the minimum distance from the origin.

Is this equivalent to say that I have to minimize $f(x,y)=x^2+y^2$ s.t. $(x − 1)^3 = y^2$ ?

Answer 1

$x-1$ cannot be negative since its cube is a square and so $x^2+y^2$ is minimum when $x=1,y=0$.

(Your understanding of what the question was asking is correct by the way.)

Answer 2

Note,

$$f(x)=x^2+y^2=x^2+(x-1)^3,\>\>\>\>\>x=1+\sqrt[3]{y^2}\ge1$$

Evaluate

$$f’(x) = 3x^2-4x+3=3(x-\frac23)^2+\frac53>0$$

Thus, $f(x)$ is strictly increasing over the domain $x\ge1$, hence $f(x)\ge f(1)=1$, where the equality represents the minimum, at $(1,0)$.