Question 717075
how can I solve this use the eccentricity of each hyperbola to find its equation in standard form center(4,1),horizontal transverse axis is 12 and eccentricity 4/3.
<pre>
You need to know 5 things about hyperbolas to solve this problem:

1. The equation of a hyperbola with horizontal transverse axis is 
{{{(x-h)^2/a^2}}}{{{""-""}}}{{{(y-k)^2/b^2}}}{{{""=""}}}{{{1}}}

2. The length of the transverse axis is {{{2a}}}, 

3. The eccentricity is {{{c/a}}},

4. The center is {{{"(h,k)"}}}

5. {{{c^2}}}{{{""=""}}}{{{a^2+b^2}}}

center(4,1) 

That tells us that h = 4 and k = 1

horizontal transverse axis is 12

That tells us that 2a = 12
                    a = 6

eccentricity 4/3

That tells us that   {{{c/a = 4/3}}}

Substituting a = 6,  {{{c/6 = 4/3}}}
Cross multiplying,   {{{3c = 24}}}
                     {{{ c = 8}}}

{{{c^2}}}{{{""=""}}}{{{a^2+b^2}}}
{{{8^2}}}{{{""=""}}}{{{6^2+b^2}}}
{{{64}}}{{{""=""}}}{{{36+b^2}}}
{{{28}}}{{{""=""}}}{{{b^2}}}

So

{{{(x-h)^2/a^2}}}{{{""-""}}}{{{(y-k)^2/b^2}}}{{{""=""}}}{{{1}}}

becomes:

{{{(x-4)^2/6^2}}}{{{""-""}}}{{{(y-1)^2/28}}}{{{""=""}}}{{{1}}}
    
{{{(x-4)^2/36}}}{{{""-""}}}{{{(y-1)^2/28}}}{{{""=""}}}{{{1}}}

Edwin</pre>