For the given curve function given as the quadratic equation:


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!

  • $\begingroup$ What about $(x+2)^2-(y+2)^2=8$ ? $\endgroup$ May 3 '19 at 23:37
  • $\begingroup$ Yep! But condidering $(x+2)^2-(y+2)^2=8$ how do I calculate the minimum aand maximum distance from the given point in an easier way? I was thinking to solve this in the following fashion: math.stackexchange.com/questions/3212658/… @SomeStrangeUser $\endgroup$
    – Cos
    May 3 '19 at 23:41
  • $\begingroup$ I'm a bit rusty when it comes calculating minimal distances from a hyperbola, but we can be sure that there is no maximum distance (as the hyperbola extends to inifinity) $\endgroup$ May 3 '19 at 23:48
  • $\begingroup$ The resulting equations don’t look particularly difficult to me, but you can simplify things a bit by translating everything so that the center of the hyperbola is at the origin (which you can read in Cos’ comment). That doesn’t affect distances. $\endgroup$
    – amd
    May 4 '19 at 0:22
  • $\begingroup$ Thanks! But can you explain a little bit further how to calculate the minimum distance this point to the hyperbola in this case? @amd $\endgroup$
    – Cos
    May 4 '19 at 0:41

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.


you can make this a one variable problem with two recipes $$ x= -2 + \sqrt 8 \cosh t \; , \; \; y = -2 + \sqrt 8 \sinh t $$ on the right branch, $$ x= -2 - \sqrt 8 \cosh t \; , \; \; y = -2 + \sqrt 8 \sinh t $$ on the left branch

I also recommend you learn how to draw the hyperbola yourself.


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