Find the points of x²-y²=16 that are closer to (0,6)

heelp

METHOD 1. Parametrically

let x = 4 sec t
y = 4 tan t

Then x²-y²=16

D^2 = (x - 0)^2 + (y - 6)^2

D^2 = 16 sec² t + (4 tan t - 6)²

2 D dD/dt = 32 tan t sec² t + 8(4 tan t - 6) sec² t

D dD/dt = sec² t (16 tan t + 16 tan t - 24)

D dD/dt = sec² t (32 tan t - 24) = 8 sec² t (4 tan t - 3)

setting dD/dt = 0, we find

8 sec² t (4 tan t - 3) = 0

sec t = 0 or tan t = 3/4

sec t = 0 has no solution

If tan t = 3/4, then sec t = ±5/4

So
x = 4 sec t = ±5
y = 4 tan t = 3

The points closest to P = (0, 6) are (-5, 3) and (5, 3).

Method 2: Implicitly
(Assume y and D depend on x.)
x² - y² = 16
2x - 2 y dy/dx = 0

dy/dx = x/y

D² = (x - 0)² + (y - 6)²

2 D dD/dx = 2x + 2(y - 6) dy/dx

Set dD/dx = 0 and substitute dy/dx = x/y

0 = 2x + 2(y - 6)(x/y)

xy + x(y - 6) = 0

2xy - 6x = 0

2x(y - 3) = 0

x = 0 or y = 3

reject x = 0

y = 3
x² = y² + 16 = 25
x = 5 or -5

Points are (5, 3) and (-5, 3)