Cho elip \((E)\) có dạng x^2 / a^2 + y^2 / b^2 =1 ( a lớn hơn b lớn hơn 0)
Giải thích
a) Đúng | b) Đúng | c) Sai | d) Đúng |
Ta có: \(\left\{ {\begin{array}{*{20}{l}}{M \in (E)}\\{N \in (E)}\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{\frac{{{5^2}}}{{{a^2}}} + \frac{{{{(\sqrt 2 )}^2}}}{{{b^2}}} = 1}\\{\frac{{{0^2}}}{{{a^2}}} + \frac{{{2^2}}}{{{b^2}}} = 1}\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{{a^2} = 50}\\{{b^2} = 4}\end{array}} \right.} \right.} \right.\). Vậy elip \((E):\frac{{{x^2}}}{{50}} + \frac{{{y^2}}}{4} = 1\).