Question

Calculus 1:
Find the points on the hyperbola y^2-x^2=4 that are closest to the point (2,0). If you decide to help, please explain the process which you obtained your answer, thanks! :)

Answer 1

I know I have to use the distance formula: \[d=\sqrt{(x-2)^2 + (y-0)^2}\left| \right| y^2=4+x^2\]
Thus I substitute in:\[y=\sqrt{(x-2)^2+x^2+4}= \sqrt{x^2-4x+4+x^2+4}\rightarrow \sqrt{2x^2-4x+8}\]

Answer 2

Wow, I see what I did wrong now...on the original paper, I didn't take the derivative...at all >.>.

Answer 3

Hey @agent0smith , post so I may metal you; I feel bad making you wait so long.

Answer 4

The site is very slow, took a while to answer. You need to minimize the distance formula. Differentiate, set equal to zero.

Answer 5

Btw DON'T differentiate the distance formula as d, differentiate d^2.

Minimize d^2 - it's much easier, and of course the minimum of d is at the same as the point as d^2. When d is a minimum, d^2 must also be/

Answer 6

a min

Answer 7

^^ on the d^2; does make EVERYTHING easier.

Answer 8

But why are we able to raise it to a power without changing what we're looking for, though? Is it just because of the simple fact that the end result will always be the derivative of the information inside the radical?

Answer 9

Because when d is a min, d^2 must be a min.

Answer 10

And yes, the end result will end up the same both ways, since the denominator (the square root) can never make the whole fraction equal zero.

Answer 11

Nice, thanks for the assistance; appreciate it. :)