How do I find the minimum distance between the line $y = 2x$ and the hyperbola $x^2 - y^2 = 3$?

Question

I've confirmed that they don't intersect, but I'm not sure where to go from there.

In addition, how can we find the points where this minimum distance occurs?

Answer 1

Find $k$ s.t. $y=2x+k$ is tangent to $x^2-y^2=3$. \begin{align} &x^2-(2x+k)^2=3\\ &3x^2+4kx+k^2+3=0\\ &D/4=4k^2-3k^2-9=k^2-9\geq0\\ &|k|\geq3 \end{align} The minimum distance ($=d$) is distance between two lines $y=2x$ and $y=2x+k$. $$d={k\over\sqrt5}={3\over\sqrt5}$$

Answer 2

The distance between two points $(x_1,y_1),(x_2,y_2)\in \mathbb R^2$ is given by

$$ \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}. $$

You want to minimize this distance subject to the points satisfying the two equations respectively. That is, you want to solve

$$ \min_{x_1,y_1,x_2,y_2} \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \quad \text{subject to} \quad y_1 = 2x_1, \quad x_2^2 - y_2^2 = 3.$$

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Solve the constraints for $y_1$ and $y_2$, you might want to split into two cases $y_2 = + \sqrt{x_2^2-3}$ and $y_2 = - \sqrt{x_2^2-3}$ but here I will do both at the same time, and plug them into the objective function to get

$$ \min_{x_1,x_2} \sqrt{(x_2-x_1)^2 + \left(\pm \sqrt{x_2^2-3}-2x_1 \right)^2} $$

with first order conditions

$$ 5x_1-x_2 = \pm 2\sqrt{x_2^2-3} , \quad \text{and} \quad 2 x_1x_2 = \pm (2x_2-x_1)\sqrt{x_2^2-3}.$$

Solving for $x_1$ and $x_2$ leads to two solutions

\begin{align} \left(\frac{4}{5},2\right)&=\arg \min_{x_1,x_2} \sqrt{(x_2-x_1)^2 + \left(+ \sqrt{x_2^2-3}-2x_1 \right)^2} \\[2ex] -\left(\frac{4}{5},2\right)&=\arg \min_{x_1,x_2} \sqrt{(x_2-x_1)^2 + \left(- \sqrt{x_2^2-3}-2x_1 \right)^2} \end{align}

and the objective function in both cases is equal to the minimal distance of $\frac{3}{\sqrt{5}}$.

Answer 3

As Hamed commented, Lagrange multipliers is the most general method for solving this kind of problems (even if, in this specific case, there are faster methods to do the work).

Just for illustration purposes, I shall use this method.

First, notice that minimizing the distance is the same as minimizing the square of the distance. So, consider the function $$F=(x_1-x_2)^2+ (y_1-y_2)^2+ \lambda(y_1-2x_1)+\mu(x_2^2-y_2^2-3)$$ Let us compute the partial derivatives which, at the minimum, will all be equal to zero. $$F'_{x_1}=2(x_1-x_2)-2\lambda$$ $$F'_{x_2}=2(x_2-x_1)+2x_2 \mu$$ $$F'_{y_1}=2(y_1-y_2)+\lambda$$ $$F'_{y_2}=2(y_2-y_1)-2y_2 \mu$$ $$F'_{\lambda}=y_1-2x_1$$ $$F'_{\mu}=x_2^2-y_2^2-3$$ From the first four equations, which are linear, we can eliminate $x_1,x_2,y_1,y_2$ and get $$x_1=\frac{\lambda (\mu +1)}{\mu }\quad \,,\quad x_2=\frac{\lambda }{\mu }\quad\,,\quad y_1=\frac{\lambda }{2 \mu }-\frac{\lambda }{2}\quad\,, \quad y_2=\frac{\lambda }{2 \mu }$$ Replacing these in the last two partial derivatives leads to $$F'_{\lambda}=-\frac{\lambda (5 \mu +3)}{2 \mu }$$ $$F'_{\mu}=\frac{3 \lambda ^2}{4 \mu ^2}-3$$ which are simple to solve; they lead to $$\mu=-\frac 35\quad\,, \quad \lambda=\pm \frac 65$$ So, for $$\lambda= +\frac 65 \implies x_1=-\frac{4}{5}\quad \,,\quad x_2=-2\quad \,,\quad y_1=-\frac{8}{5}\quad \,,\quad y_2=-1\quad \,,\quad d^2=\frac{9}{5}$$ for $$\lambda= -\frac 65 \implies x_1=+\frac{4}{5}\quad \,,\quad x_2=+2\quad \,,\quad y_1=+\frac{8}{5}\quad \,,\quad y_2=+1\quad \,,\quad d^2=\frac{9}{5}$$

For sure, the problem could have been made simpler because one of the constraints is stritly linear; so we could have used instead $$F=(x_1-x_2)^2+ (2x_1-y_2)^2+ \mu(x_2^2-y_2^2-3)$$ and proceed the same way.

Making the problem more general for $y=ax$ and $x^2-y^2=b$, the same method would lead to $$d^2=\frac{\left(a^2-1\right) b}{a^2+1}$$