Question

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?

Answer 1

`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(e^2-1). 19 = 4 ( e^2-1)`,

So, the eccentricity of the hyperbola `e = sqrt 23 / 2`.

The least distance of the comet from the Sun

= the distance between the branch vertex and its focus

`= a(e-1)=2(sqrt23 / 2- 1)=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 `oo`.