Question

Find the point(s) (x,y) on the hyperbola y^2 = 1+(x^2/4) that is are closest to the point (6,0). Calculus. Please explain how to work the problem.

Answer 1

IDK sorry

Answer 2

idk i will try to figure out though

Answer 3

\[y^2= 1+\frac{ x^2 }{ 4 }\] just to clarify the equation and thank you for the help

Answer 4

well you need to get y alone by square rooting the problem for starters

Answer 5

that's one way of doing it. I personally think it'd be easier just to derive it implicitly.

Answer 6

yes that is one way

Answer 7

so \[y'2y = \frac{ x }{ 2 }\]

therefore
\[y' = \frac{ x }{ 4y }\]

Answer 8

you are a guiness

Answer 9

this is where I'm lost though. I don't know how to find out the distance. I can get the critical points here but at (6,0) this derivative is infinity which shows the point clearly doesn't exist, which is fine. but I'm just lost. I'll look it up online or somethin

Answer 10

im a fricken 8 year old i don't know anything!

Answer 11

sike I'm actually 17:)

Answer 12

are you really?

Answer 13

oh okay xD I'm only 18 so its cool

Answer 14

I would define a function that represents the distance from the point (6,0) to the hyperbola, and then minimize that function.

It is perhaps easier to deal with the square of the distance (minimizing the distance squared will also minimize the distance)

use the distance formula
\[ D^2 = (x_1 - x_0)^2 + (y_1 - y_0)^2 \]
use (6,0) for \( (x_0, y_0) \)
and
\[ y = \sqrt{1 + \frac{x^2}{4}} \]

Answer 15

we have the problem of minimizing the function
\[ f(x) = (x - 6)^2 + \left(0 - \sqrt{1+\frac{x^2}{4} }\right)^2\\
f(x) = (x - 6)^2 + 1 + \frac{x^2}{4}
\]

Answer 16

Here is a graph of the problem