Question

Answer 1

Theo đề bài ta có:

\(\left\{{}\begin{matrix}2c=8\\\dfrac{b}{a}=\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=4\\a=3b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^2-b^2=4\\a^2=9b^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^2=\dfrac{9}{2}\\b^2=\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left(E\right):\dfrac{x^2}{\dfrac{9}{2}}+\dfrac{y^2}{\dfrac{1}{2}}=1\)

Answer 2

Giả sử a > b > 0 ; a > c > 0 . Ta có : \(\dfrac{b}{a}=\dfrac{1}{3};2c=8\)

Suy ra : \(a=3b;c=4\)

Mặt khác : \(c=\sqrt{a^2-b^2}=\sqrt{\left(3b\right)^2-b^2}=2\sqrt{2}b=4\)

\(\Rightarrow b=\sqrt{2}\) \(\Rightarrow a=3\sqrt{2}\)

PTCT của elip : \(\dfrac{x^2}{18}+\dfrac{y^2}{2}=1\)