Cho parabol có phương trình \({y^2}\; = 12x.\)

Tìm toạ độ tiêu điểm của parabol.

\({c^2} = {a^2} + {b^2}\).    

\({b^2} = {a^2} + {c^2}\).

\({a^2} = {b^2} + {c^2}\).

\(c = a + b\) .

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

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

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

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

A.\(46,92\) cm

B.\(23,26\) cm

C.\(65\) cm

D.\(11,63\) cm

\(2a = {F_1}{F_2}\). 

\(2a > {F_1}{F_2}\).

\(2a < {F_1}{F_2}\).

\(4a = {F_1}{F_2}\).

$6$

$2\sqrt {41} $

$3$     

$\sqrt {41} $

x = 3.

x = –3.

y = –3.

y = 3.