Closest point to $(2,0)$ on with a hyperbola as a constraint

Question

I'm looking to find a point on the hyperbola $y^{2}-x^{2}=4$ which is closest to $(2,0)$.

As far as I know I need to find the distance formula and use lagrange multipliers.

Answer 1

the formula of distance is $$d^2=(x-x_0)^2+(y-y_0)^2$$ $$d^2=(x-2)^2+(\sqrt{4+x^2}-0)^2 $$ $$d^2=(x-2)^2+(4+x^2)$$ now we assume the $d^2$ $=f(x)$ $$f(x)=(x-2)^2+(4+x^2)$$ $$f'(x)=2(x-2)+2x=0$$ $$x=1$$ $$y=\pm \sqrt{5}$$

Answer 2

I do not think you need LaGrange multiplier. Define $f(x)$ to be the square of the distance between the point $(2,0)$ and the hyperbola. (It is enough to minimize the square of the distance to make the calculus easier). Using the top half of the hyperbola, with a bit of math we get $$f(x)=(x-2)^2+(\sqrt{x^2+4}-0)^2.$$ Minimize this function using the first derivative test to find the value of $x$ on the hyperbola closest to $(2,0).$ Then use the hyperbola equation to find the $y-$value.

Answer 3

HINT: use $d=\sqrt{(2-x)^2+4+x^2}$

Answer 4

minimize $(x-2)^2 + y^2$ subject to $y^2 - x^2 = 4$

using Lagrange multipliers you will end up with:
$2(x-2) + \lambda(-2x) = 0$
$2y + \lambda(2y) = 0$
$y^2 - x^2 = 4$

you get $\lambda = -1$, $x=1$, and $y=\sqrt{5},-\sqrt{5}$