# Finding minimum and maximum distance from a tricky curve equation to a given point without Lagrange multipliers.

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!

• What about $(x+2)^2-(y+2)^2=8$ ? – SomeStrangeUser May 3 '19 at 23:37
• 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 – Cos May 3 '19 at 23:41
• 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) – SomeStrangeUser May 3 '19 at 23:48
• 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. – amd May 4 '19 at 0:22
• Thanks! But can you explain a little bit further how to calculate the minimum distance this point to the hyperbola in this case? @amd – 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.