# Shortest distance between an ellipse and a hyperbola.

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?

• I don’t know if this is going to end up being any easier, but you can eliminate one parameter by finding the points on the hyperbola at which the tangents are parallel to the ellipse’s tangent at $(a\cos\theta,b\sin\theta)$. – amd Jun 17 '17 at 2:11
• – lab bhattacharjee Jun 17 '17 at 2:27

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. 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 $PQ=a(\sqrt m-1)$.

• n i c e .. .... – janmarqz Apr 10 at 18:36
• @janmarqz - t h a n k s ! – hypergeometric Apr 12 at 16:32

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.