For the given curve function given as the quadratic equation:
$$g(x,y)=x^2+4x-y^2-4y-8=0;$$
and $P=(8,6) \in \mathbb{R}^{2}$. Im finding the minimum and maximum distance from $P$ to the curve $g(x,y)$. I can solve this using Lagrange multipliers method by taking $$f(x,y)=(x-8)^2+(y-6)^2,$$
the squared distance from $P$ and finding $x,y$ and $\lambda$ such $\nabla f= \lambda\nabla g$. But for this particular is hard to find $\lambda$ using Lagrange multipliers method. Im almost sure this problem can be solved a lot easier, so I was thinking to factor $g(x,y)$ as a more common curve but I cant do it. I meant to factor $g(x,y)$ as $$(x+2)^2+(y-2)^2=16$$ but this is not correct as I originally have $-y^2$. Any help finishing this proof the easier way will be appreciated. Thanks!
As the equation can be written as $(x + 2)^2 - (y + 2)^2 = 8$, we can simplify it by doing a translation $x' = x + 2$ and $y' = y + 2$
For the ease of typing the variable x, y will be used instead of $x'$, $y'$.Then the question becomes finding the minimum distance between the curve $x^2 - y^2 = 8$ and the point $(10, 8)$.
Since the distance should be the perpendicular distance from the point to the curve, let any point $(a, b)$ on the curve, the normal at this point will be $$\frac{y - b}{x - a} = - \frac{b}{a}$$
Since the point $(10, 8)$ passes through this normal, we have
$$\frac{8 - b}{10 - a} = -\frac{b}{a} \implies b = \frac{4a}{a - 5}$$
Also $(a, b)$ is on the curve, $a^2 - b^2 = 8$. Solving these 2 equations we have
$$a^4 - 10a^3 + a^2 + 80a - 200= 0$$
By trial and error we have
$a = 9.20$ and $b = 8.76$ giving distance = 1.10
Another solution is
$a = - 3.24$ and $b = 1.57$ giving distance = 14.72
Then the minimum distance is 1.10.