If there is a point (a,b) on the hyperbola $x^2-y^2=4$ that is closest to the point (0,1), then what is this point?
I prefer this problem to be solved with the knowledge of the the maximum and minimum using the derivatives and so on.
I am a bit confused
I said that I will take the square distance between the point $0,1$ and $y=$$\sqrt{x^2-4}$
then I got $D(x)=x^2-(\sqrt{x^2-4}-1)^2$
after that I took the derivative and made it equal to zero and I was lost at this point since I didn't know how to get the values that would lead me to the Minimum.
?!?!
Best Answer
It's probably easier to take $x=\pm\sqrt{y^2+4}$, since this is true for all $y$, while looking at the question in terms of $x$ requires the condition that $|x|\geq 4$.
Also, $x^2+(y-1)^2 = y^2+4 +(y-1)^2$ is much easier to minimize, and doesn't depend on the sign of $x$.