suppose $(x_0,y_0)$ is a point on some curve $g(t):R->R_2$ that is Closest to a point $(x_1,y_1)$ NOT on the curve .
show that the line through $(x_0,y_0)$ and $(x_1,y_1)$ is orthogonal to the tangent line to the curve at the point $(x_0,y_0)$
really though question i have tried everything
let $g(t) = (x(t) , y(t))$
$g'(t) = (x'(t),y'(t))$
so we can say that the tangent line $(x_0,y_0) + s(x',y')$ and the line that passes through the points : $(x_0,y_0)+k(x_1-x_0,y_1-y_0)$
and we need to prove that those lines orthogonals , i tried dot product equals to 0 but i always find that i miss something to prove.
i am sure that this missing clue is that we need to use the fact that this 2 points are the closest to each other and that no other point ON THE CURVE is closer.
Your attempted solution is incorrect. Firstly, the closest-point condition is not sufficient to imply that the curve is differentiable there. So the question actually must have that extra condition. Secondly, you wrongly assumed that the curve is a graph of $(x,y)$ where $y$ is some function of $x$. That was not given. $ \def\lfrac#1#2{{\large\frac{#1}{#2}}} $
Let me use variables $x,y$ instead of $x_0,y_0$, varying with parameter $t$. We can also assume that $x_1 = y_1 = 0$.
Let $r = x^2+y^2$. Consider any point on the curve (specified by parameter $t$ and coordinates $x,y$). As $Δt \to 0$ we have $\lfrac{Δr}{Δt} = \lfrac{((x+Δx)^2+(y+Δy)^2) - (x^2+y^2)}{Δt}$ $= 2(x\lfrac{Δx}{Δt}+y\lfrac{Δy}{Δt}) + (Δx\lfrac{Δx}{Δt}+Δy\lfrac{Δy}{Δt})$. If the curve is differentiable there, then we also have $Δx,Δy \approx 0$ and $\lfrac{Δx}{Δt} \approx \lfrac{dx}{dt}$ and $\lfrac{Δy}{Δt} \approx \lfrac{dy}{dt}$, and so $\lfrac{Δr}{Δt} \approx 2(x\lfrac{dx}{dt}+y\lfrac{dy}{dt})$, and hence $\lfrac{dr}{dt} = 2(x\lfrac{dx}{dt}+y\lfrac{dy}{dt})$. If moreover $r$ is minimum there, then $\lfrac{dr}{dt} = 0$, and so $x\lfrac{dx}{dt}+y\lfrac{dy}{dt} = 0$, equivalently $(x,y)·(\lfrac{dx}{dt},\lfrac{dy}{dt}) = 0$, and hence $(x,y)$ and $(\lfrac{dx}{dt},\lfrac{dy}{dt})$ are orthogonal.
Note that $(x,y)$ and $(\lfrac{dx}{dt},\lfrac{dy}{dt})$ are orthogonal if and only if they are perpendicular or one of them is zero. So there is actually no need for the question to state that $(x_1,y_1)$ is not on the curve. If you really want to get perpendicularity, even that condition is insufficient; you also have to add an extra condition that the curve's derivative at that point is nonzero. Otherwise it is possible that $(\lfrac{dx}{dt},\lfrac{dy}{dt})$ is zero (the point tracing the curve stops momentarily) but there is no (geometric) tangent there; for example $(\cos(t)+\sin(t),\sin(2t))$ has zero derivative when $t = \lfrac14π$.