Question 1198028
given:

two stations on the shore {{{168}}} miles apart that are located at the {{{foci}}}


so, the coordinates of focus will be F({{{84}}},{{{0}}}) and F({{{-84}}},{{{0}}}) => {{{c=84}}}

enter is at origin =>{{{h=0 }}} and {{{k=0}}}

 the ship is {{{60}}} miles south of the center of the hyperbola=> {{{b=60

then

{{{c^2=a^2+b^2}}}

{{{84^2=a^2+60^2}}}

{{{a^2=84^2-60^2}}}

{{{a^2=3456}}}

{{{a=sqrt(3456)}}}

{{{a=24sqrt(6)}}}



The general equation of hyperbola you need is:

{{{x^2/a^2-y^2/b^2=1}}}

So, the equation of path of ship which is in the form of hyperbola is

{{{x^2/(24sqrt(6))^2-y^2/60^2=1}}}

{{{x^2/3456-y^2/3600=1}}}


{{{ drawing( 600, 600, -150, 150, -150, 150,
circle(-84,0,2),circle(84,0,2),
locate(-84,8,F(-84,0)),locate(84,8,F(84,0)),
graph( 600, 600, -150, 150, -150, 150,-(5sqrt(x^2-3456))/(2sqrt(6)),  (5sqrt(x^2-3456))/(2sqrt(6)))) }}}