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Suppose we have a hyperbola $xy=c$ $(c\neq0)$ and an ellipse $x^2/a^2+y^2/b^2=1$ $(a>b>0)$ that do not intersect. What is a quick way to calculate the shortest distance between these two curves?

I thought of setting a point $(a\cos\theta,b\sin\theta)$ on the ellipse and a point $(t,c/t)$ on the hyperbola, then the distance is given by $$\sqrt{(a\cos\theta-t)^2+(b\sin\theta-\frac{c}{t})^2},$$ whose minimum may be found by calculus methods. But the computations look tedious and difficult to carry out. What are some easier ways to do this?

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There doesn't seem to be a nice closed-form solution to this, but the following might be helpful..

For an ellipse $\frac {x^2}{a^2}+\frac {y^2}{a^2}=1$, the equation of the normal at point $P(a \cos\theta, b\sin\theta)$ on the ellipse is

$$\frac {a\sin\theta}{b\cos\theta}x-y=a\left(\frac{a^2-b^2}{ab}\right)\sin\theta\tag{1}$$

A hyperbola which just touches or does not intersect at all with the ellipse above has the equation $xy=\frac {mab}2$ where $m\ge 1$. The equation of the normal at point $Q (v,\frac {mab}{2v})$ on the hyperbola is $$\frac {2v^2}{mab}x-y=v\left(\frac {2v^2}{mab}-\frac {mab}{2v^2}\right)\tag{2}$$

The minimum distance between the ellipse and hyperbola is the distance $PQ$ when $(1)=(2)$, i.e. both normals are the same line.

As the coefficients of $y$ are the same in both $(1),(2)$, equating coefficients of $x$ in $(1),(2)$ gives $$v=a\sqrt \frac{m\tan\theta}{2}\tag{3}$$ This relationship ensures that the tangents and normals at $P,Q$ are parallel to each other respectively (but the normals are not necessarily the same line).

Putting $(3)$ in $(2)$ gives

$$\left(\frac ab \tan\theta\right)x-y=a\sqrt{\frac{m\tan\theta} 2}\left(\frac ab\tan\theta-\frac ba\cot\theta\right)\tag{4}$$

To ensure that both normals are the same line, we need to equate RHS of $(1),(4)$. This gives $$\left(\frac{a^2-b^2}{ab}\right)\sin\theta=\sqrt{\frac{m\tan\theta}2}\left(\frac ab \tan\theta-\frac ba\cot\theta\right)$$ which is equivalent to $$(a^2-b^2)\sin\theta=\sqrt{\frac{m\tan\theta}2}\left(\frac{a^2\sin^2\theta-b^2\cos^2\theta}{\sin\theta\cos\theta}\right)\tag{4}$$

Solve numerically $\theta$ in $(4)$, find corresponding value $v$ in $(3)$, then calculate $PQ$. This should give the minimum distance between the ellipse and hyperbola.

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See desmos implementation here.


In the trivial case where $a=b$ (i.e. ellipse is a circle), then $\theta=\frac \pi 4$ and $v=a\sqrt{\frac m2}$ . This gives $P=\left(\frac a{\sqrt{2}}, \frac a{\sqrt{2}}\right)$ and $Q=\left(a\sqrt{\frac m2}, a\sqrt{\frac m2}\right)$ and the distance $PQ=a(\sqrt m-1)$.

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  • $\begingroup$ n i c e .. .... $\endgroup$ – janmarqz Apr 10 at 18:36
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    $\begingroup$ @janmarqz - t h a n k s ! $\endgroup$ – hypergeometric Apr 12 at 16:32
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There's easy draft solution to this problem:

  1. First scale the ellipse so that it becomes a circle -- this allows calculating the distance from the "circle", since $|d(p)-r|$ is the distance.
  2. Then build a distance space, $f :: (x,y) \rightarrow (d)$ which contains the distance values to the circle. $f(x,y) = |\sqrt{x^2+y^2}-r|$.
  3. Then create $g :: (x) \rightarrow (x_0,y_0)$ from the hyperbola via $g(x) = (x,c/x)$
  4. Use scaling function $s :: (x_1,y_1) \rightarrow (x_2,y_2)$ which matches the scaling used in step (1).
  5. Use function composition of the three funtions to get $f \circ s \circ g :: (x) \rightarrow (d)$.
  6. Use derivative $= 0$ to find point where (d) is smallest.

Most difficult part is getting the scaling right, since values of (d) depends on how the scaling is being done.

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