# A comet follows the hyperbolic path described by x^2/4 -y^2/19 = 1, where x and y are in millions of miles. If the sun is the focus of the path, how close to the sun is the vertex of the path?

May 11, 2016

$\sqrt{23} - 2 = 2.796$ million miles, nearly.,

#### Explanation:

Unit of distance is 1 million miles.

Semi-transverse axis a = 2, semi-conjugate axis$b = \sqrt{19}$.

${b}^{2} = {a}^{2} \left({e}^{2} - 1\right) . 19 = 4 \left({e}^{2} - 1\right)$,

So, the eccentricity of the hyperbola $e = \frac{\sqrt{23}}{2}$.

The least distance of the comet from the Sun

= the distance between the branch vertex and its focus

$= a \left(e - 1\right) = 2 \left(\frac{\sqrt{23}}{2} - 1\right) = \sqrt{23} - 2 = 2.796$ million miles, nearly.

The path of a comet might be nearly parabolic, with eccentricity $1 -$, like e = 0.95. Possibly, somewhere in the orbit, it travels, piece-wise, along a hyperbola.

The whole path is unlikely to be a hyperbola, with orbital period $\infty$.