Distance of a Point from Hyperbola

Question

Consider the part of hyperbola $H_{+}=\{(x,1/x)\colon x>0\}$ in the first quadrant, and $(a,b)$ any point in the plane (for sake of convenience, say $a,b>0$). If $(a,b)$ does not lie on the hyperbola $H_{+}$, how to determine

(1) The (minimum) distance of $(a,b)$ from $H_{+}$

(2) The point of $H_{+}$ which is closest to $(a,b)$.

I would like to determine these things algebraically as well as geometrically.

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The way I proceeded is as follows: if $(y,1/y)$ is closest to $(a,b)$, then $y$ satisfies a fourth degree equation of the type $Y^3(Y-a)=(1-bY)$. But I couldn't proceed further.

Answer 1

Let $(y,1/y) \in H_+$ be the point in $H_+$ closest to $(a,b)$.

The gradient of $H_+$ at $(y, 1/y)$ is given by $\frac{d}{dx}(1/x) = -\frac{1}{x^2}$.

So the tangent to $H_+$ at $(y,1/y)$ is the set $\{(y,1/y) + r(-1/y^2,1):r \in \mathbb{R}\}$.

This tangent is perpendicular to the straight line joining $(y,1/y)$ and $(a,b)$, so we have

$$ \left( \begin{array}{c} a-y\\ b-1/y\\ \end{array} \right) \cdot \left( \begin{array}{c} -1/y^2\\ 1\\ \end{array} \right) = 0. $$

The positive solution to this quadratic is $y = \sqrt{\frac{a}{b}}$, so the closest point to $H_+$ is $\left(\sqrt{\frac{a}{b}}, \frac{1}{\sqrt{\frac{a}{b}}}\right)$, and this can be used to work out the shortest distance from $(a,b)$ to $H_+$.

Answer 2

Let the curve $y=1/x$ be parametrized by $P(t)=(x,y)=(t,1/t).$ Then the squared distance from $(a,b)$ to $P(t)$ has derivative $(2/t^2)[t^4-at^3+bt-1],$ and so ignoring the $2/t^2$ factor the critical points are the roots of $$g(t)=t^4-at^3+bt-1,$$ just as stated in the posted question (with the letter $Y$ used for $t$).This would be a stopping point in general, except for numerical methods given specific $a,b,$ if one wants to avoid solving a general quartic equation.

An interesting (very) special case is that in which $a=b$ [call each $k$] with $(k,k)$ in quadrant I and lying say above the curve $y=1/x.$ Here loking at it geometrically we should expect the two points $(1,1)$ and $(-1,-1)$ on the full hyperbola to give critical points, so that $g(t)$ should factor leaving a quadratic. Indeed with $a=b=k$ we get to $g(t)=(t-1)(t+1)(t^2-kt+1).$ The quadratic here has no real roots when $k<2,$ while if $k=2$ it is $(t-1)^2$ making an order 3 zero for $g(t)$ at $t=1.$ What seems interesting (to me) is that when $k>2$ the two values of $t$ from the quadratic lead to two points on $y=1/x$ which win out over the point $(1,1)$ as being closer to it on the curve. The squared distance from $(k,k)$ to $(1,1)$ is $2(1-k)^2,$ while after some algebra the squared distance from either of the other two points (from the quadratic) is $k^2-2.$ And if one subtracts this from $2(1-k)^2$ the result is $(k-2)^2>0.$ [note we need $k \ge 2$ anyway for this expression to refer to a real difference of squared lengths in the situation of the problem]

Answer 3

To add onto solving @coffeemath's polynomial equality

$$x^4-ax^3+bx-1=0,$$ we begin by re-writing the polynomial as follows:

$$(x^2+Ax+B)^2=Cx^2+Dx+E.$$ Now expand and equate coefficients, adding the extra condition $D^2-4CE=0.$ Once done, take square roots,

$$x^2+Ax+B=\pm_1\sqrt{C}\left(x+\frac{D}{2C}\right)$$ and solve the quadratic.