Question 861944
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Hi
{{{x2/169-y2/16=1 }}}
{{{x2/13^2-y2/4^2=1 }}}
  a = 13  and b = 4  slopes  m =  ± 4/13
Standard Form of an Equation of an Hyperbola opening right and  left is:
  {{{(x-h)^2/a^2 - (y-k)^2/b^2 = 1}}} with C(h,k) and vertices 'a' units right and left of center,   2a the length of the transverse axis
Foci are {{{sqrt(a^2+b^2)}}} units right and left of center along y = k
& Asymptotes Lines passing thru C(h,k), with slopes  m =  ± b/a 
{{{drawing(300,300, -10, 10, -10, 10,  grid(1),
graph( 300, 300,  -10, 10, -10, 10,0, (4/13)x, (-4/13)x, 4sqrt(x^2/169 +1), -4sqrt(x^2/169 + 1)))}}}