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?

1 Answer
May 11, 2016

#sqrt 23 - 2=2.796# million miles, nearly.,


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#.