Phương trình chính tắc của elip có đi qua hai điểm \[M\left( {2\sqrt 2 ;\frac{1}{3}} \right)\] và \[N\left( {2;\frac{{\sqrt 5 }}{3}} \right)\] là:

A.\[{y_M} \in \left( {0;\sqrt 3 } \right)\]

B. \[{y_M} \in \left( {2;\sqrt 8 } \right)\]

C. \[{y_M} \in \left( {\sqrt 8 ;5} \right)\]

D. \[{y_M} \in \left( {\sqrt 3 ;2} \right)\]

A.\[\frac{{{x^2}}}{{36}} + \frac{{{y^2}}}{{16}} = 1\]

B. \[\frac{{{x^2}}}{{144}} + \frac{{{y^2}}}{{64}} = 1\]

C. \[\frac{{{x^2}}}{{12}} + \frac{{{y^2}}}{8} = 1\]

D. \[\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{36}} = 1\]

A.\[\frac{{{x^2}}}{{24}} + \frac{{{y^2}}}{{25}} = 1\]

B. \[\frac{{{x^2}}}{{24}} + \frac{{{y^2}}}{{25}} = - 1\]

C. \[\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{{24}} = 1\]

D. \[\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{{24}} = - 1\]Trả lời:

A.\[\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{{16}} = 1\]

B. \[\frac{{{x^2}}}{5} + \frac{{{y^2}}}{4} = 1\]

C. \[\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{9} = 1\]

D. \[\frac{x}{5} + \frac{y}{4} = 1\]

A.\[{c^2} = {a^2} + {b^2}\]

b. \[{b^2} = {a^2} + {c^2}\]

C. \[{a^2} = {b^2} + {c^2}\]

D. \[c = a + b\]

A.\[2a = {F_1}{F_2}\]

B. \[2a >{F_1}{F_2}\]

C. \[2a < {F_1}{F_2}\]

D. \[4a = {F_1}{F_2}\]