Question 1198410

 The hyperbola in the tower can be modeled by 

{{{x^2/25 - y^2/256 =1 }}}

with {{{150m}}} as its height. 


<a href="https://ibb.co/QDMk3LH"><img src="https://i.ibb.co/QDMk3LH/Capture11-20-2022-5-05-35-PM.jpg" alt="Capture11-20-2022-5-05-35-PM" border="0"></a>


center is at ({{{0}}},{{{ 0}}})

{{{a=5}}}
{{{b=16}}}
semimajor axis length | {{{5}}}
semiminor axis length | {{{16}}}

vertices | ({{{-5}}}, {{{0}}}) | ({{{5}}}, {{{0}}})


the narrowest part  in the middle is the  distance between   vertices or 
{{{2a=2*5=10m}}}

 the  width on the top is  the  distance between points  ({{{-x}}}, {{{75}}}) and ({{{x}}}, {{{75}}})

now  find {{{x }}}  when {{{y=75}}}

{{{x^2/25 - 75^2/256 =1 }}}

{{{x = (5 sqrt(5881))/16=23.97}}}

or

{{{x = (5 sqrt(5881))/16=-23.97}}}


*[invoke formula_distance -23.97, 75, 23.97, 75] 


so the  width on the top  is:  {{{47.94m}}}



answer: 

 the width of the tower at the top is {{{47.94m}}}
and 
its narrowest part in the middle is {{{10m}}}