Đáp án:
$\begin{array}{l}
a)\left\{ \begin{array}{l}
2a = 8\\
2b = 6
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a = 4\\
b = 3
\end{array} \right.\\
\Leftrightarrow \left( E \right):\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\\
hay\,\left( E \right):\dfrac{{{x^2}}}{{16}} + \dfrac{{{y^2}}}{9} = 1
\end{array}$
$\begin{array}{l}
b)\left\{ \begin{array}{l}
2a = 10\\
2c = 6
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a = 5\\
c = 3
\end{array} \right.\\
\Leftrightarrow {b^2} = {a^2} - {c^2} = {5^2} - {3^2} = 16\\
\Leftrightarrow \left( E \right):\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\\
hay\,\left( E \right):\dfrac{{{x^2}}}{{25}} + \dfrac{{{y^2}}}{{16}} = 1
\end{array}$
