mình sửa bài nhé
b, \(c=\sqrt{a^2-b^2}=\dfrac{5\sqrt{3}}{2}\)
\(F1\left(-\dfrac{5\sqrt{3}}{2};0\right);F2\left(\dfrac{5\sqrt{3}}{2};0\right)\)
\(F1F2=2c=5\sqrt{3}\)
a, cho pt chính tắc (E) có dạng \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\)
Ta có hệ \(\left\{{}\begin{matrix}\dfrac{25}{a^2}=1\\\dfrac{5}{a^2}+\dfrac{5}{b^2}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^2=25\\b^2=\dfrac{25}{4}\end{matrix}\right.\)
-> \(\dfrac{x^2}{25}+\dfrac{y^2}{\dfrac{25}{4}}=1\)
b, \(c^2=a^2+b^2\Rightarrow c=\dfrac{5\sqrt{5}}{2}\)
\(F1\left(-\dfrac{5\sqrt{5}}{2};0\right);F2\left(\dfrac{5\sqrt{5}}{2};0\right)\)
\(2c=5\sqrt{5}\)
