Tọa độ các tiêu điểm của hypebol $(H):\,\,{x^2} - {y^2} = 1$ là:

${A_1}\left( { - 16;0} \right);\,\,{A_2}\left( {16;0} \right);\,\,{B_1}\left( {0; - 9} \right);\,\,{B_2}\left( {0;9} \right)$

${A_1}\left( { - 4;0} \right);\,\,{A_2}\left( {4;0} \right);\,\,{B_1}\left( {0; - 3} \right);\,\,{B_2}\left( {0;3} \right)$

${A_1}\left( { - 4;0} \right);\,\,{A_2}\left( {4;0} \right);\,\,{B_1}\left( {0; - 9} \right);\,\,{B_2}\left( {0;9} \right)$           

${A_1}\left( { - 16;0} \right);\,\,{A_2}\left( {16;0} \right);\,\,{B_1}\left( {0; - 3} \right);\,\,{B_2}\left( {0;3} \right)$

$2;\,\,4$

$4;\,\,2$

$2\sqrt 2 ;\,\,4$

$4;\,\,2\sqrt 2$

$6$

$2\sqrt {41}$

$3$     

$\sqrt {41}$

$y = \dfrac{3}{4}x;\,\,\,y =  - \dfrac{3}{4}x$

$y = \dfrac{4}{3}x;\,\,y =  - \dfrac{4}{3}x$

$y = \dfrac{9}{{16}}x;\,\,\,y =  - \dfrac{9}{{16}}x$         

$y = \dfrac{{16}}{9}x;\,\,\,\,y =  - \dfrac{{16}}{9}x$

$e = \dfrac{5}{4}$

$e = \dfrac{4}{5}$                

$e = \dfrac{3}{4}$    

$e = \dfrac{4}{3}$

$\dfrac{{{x^2}}}{{16}} - \dfrac{{{y^2}}}{{25}} = 1$.

$\dfrac{{{x^2}}}{{25}} - \dfrac{{{y^2}}}{9} = 1$.

$\dfrac{{{x^2}}}{9} - \dfrac{{{y^2}}}{{25}} = 1$.        

$\dfrac{{{x^2}}}{{16}} - \dfrac{{{y^2}}}{9} = 1$.