Find all points on the curve x^2-y^2=1 closest to (0,2)

Question

I know i need to minimize the distance, but I am having trouble figuring out what equations or functions to use.

Answer 1

If you consider the graph of x² - y² = 1 is a hyperbola and if you superimpose the point (0, 2) you get something like this:

Obviously, the only portion of the graph that we need to worry about is the upper portion (above the x-axis). That portion of the graph can be written by solving the graph for y and only taking the positive part.

x² - y² = 1

x² - 1 = y²

y = ± √(x² - 1) but, taking only the positive part we get y = √(x² - 1)

The distance formula is d = √[(x₂ - x₁)² + (y₂ - y₁)²] and let's let (x₂, y₂) be the point (0, 2) so the distance formula becomes d = √[(0 - x₁)² + (2 - y₁)²]

Now let's just drop the "1" suffix from the x and y to make it easier and simplify to get: d = √[x² + 4 - 4y + y²] where (x, y) is on the graph of x² - y² = 1

Now, we know that y = √(x² - 1) so we can plug "√(x² - 1)" into the equation wherever we see "y" making the distance equation: d = √[x² + 4 - 4√(x² - 1) + (√(x² - 1))²]

If we simplify and then write as a function of x, we get:

d(x) = √(2x² - 4√(x² - 1) + 3) or d(x) = [2x² - 4(x² - 1)^1/2 + 3]^1/2

To minimize, take the derivative and set it equal to zero

d '(x) = 1/2(2x² - 4√(x² - 1) + 3)^-1/2 • (4x - 2(x² - 1)^-1/2 •2x)

The only way this can be equal to zero is if the numerator equals zero so:

4x - 4x/(x²-1)^1/2 = 0

4x(1 - 1/(x²-1)^1/2) = 0 so x = 0 or 1 - 1/(x²-1)^1/2) = 0

1 = 1/(x²-1)^1/2

(x²-1)^1/2 = 1

x² - 1 = 1

x² = 2

x = ±√2

When x = √2, then y = √((√2)² - 1) = √(2 - 1) = √1 = 1 so the point is (√2, 1). The corresponding point on the other side is (-√2, 1)

Answer 2

Pick an arbitrary point (x,y) on the given curve. Then the distance from (x,y) to (0,2) is d= sqrt(x² + (y-2)²). Now you want to minimize d. But you can also minimize d^2 instead. Do you see why????

So d² = x² + (y-2)^2. Get d² in terms of one variable and then find the minimum value.