Question

Encontrar sobre a curva x²-y²=1 o ponto mais proximo o ponto (0,2)

Answer 1

distancia entre os pontos (x,y) e (x0, y0)

$D^2= (x-x_0)^2+(y-y_0)^2$

temos
x0, y0 = (0,2)

x²-y²= 1
x² = 1+y²

substituindo
$D^2=(x-0)^2+(y-2)^2\\\\d^2=x^2+(y^2-2y+4)\\\\D^2=x^2+y^2-2y+4\\\\D^2=(1+y^2)+y^2-2y+4\\\\\boxed{D^2=2y^2-2y+5}$

a menor distancia vai ser quando a derivada da distancia for igual a 0
dD/dy =0

$2D* \frac{dD}{dy} = 4y-2+0\\\\0= 4y-2\\\\\boxed{\boxed{y= \frac{1}{2} }}$

como
$x^2-y^2=1\\\\x^2- (\frac{1}{2})^2=1\\\\x^2=1+ \frac{1}{4} \\\\x=\pm \frac{\sqrt{5}}{2}$

ponto 
$P=\left( \frac{\sqrt{5}}{2} , \frac{1}{2} \rigjt)$