ta có tâm sai là $\dfrac{c}{a}=\dfrac{2}{3}$ $\Rightarrow c=\dfrac{2a}{3}\\ c^2=a^2-b^2\\ \Leftrightarrow \left (\dfrac{2a}{3} \right )^2=a^2-b^2\\ \Rightarrow b^2=a^2-\dfrac{4a^2}{9}=\dfrac{5a^2}{9}$ Phương trình elip có dạng $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=14$ Do $A(1;2)\in (E)\Rightarrow \dfrac{1^2}{a^2}+\dfrac{2^2}{\dfrac{5a^2}{9}}=1$ $\Leftrightarrow \dfrac{1}{a^2}+\dfrac{4.9}{5a^2}=1\\ \Leftrightarrow 5+36=5a^2\\ \Leftrightarrow 5a^2=41\\ \Leftrightarrow a^2=\dfrac{41}{5}\\ \Rightarrow b^2=\dfrac{5.\dfrac{41}{5}}{9}=\dfrac{41}{9}$ Phương trình elip có dạng $\dfrac{x^2}{\dfrac{41}{5}}+\dfrac{y^2}{\dfrac{41}{9}}=1$
Đáp án:
$\dfrac{x^2}{\dfrac{41}{5}}+\dfrac{y^2}{\dfrac{41}{9}}=1$
Giải thích các bước giải:
ta có tâm sai là $\dfrac{c}{a}=\dfrac{2}{3}$
$\Rightarrow c=\dfrac{2a}{3}\\
c^2=a^2-b^2\\
\Leftrightarrow \left (\dfrac{2a}{3} \right )^2=a^2-b^2\\
\Rightarrow b^2=a^2-\dfrac{4a^2}{9}=\dfrac{5a^2}{9}$
Phương trình elip có dạng $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=14$
Do $A(1;2)\in (E)\Rightarrow \dfrac{1^2}{a^2}+\dfrac{2^2}{\dfrac{5a^2}{9}}=1$
$\Leftrightarrow \dfrac{1}{a^2}+\dfrac{4.9}{5a^2}=1\\
\Leftrightarrow 5+36=5a^2\\
\Leftrightarrow 5a^2=41\\
\Leftrightarrow a^2=\dfrac{41}{5}\\
\Rightarrow b^2=\dfrac{5.\dfrac{41}{5}}{9}=\dfrac{41}{9}$
Phương trình elip có dạng $\dfrac{x^2}{\dfrac{41}{5}}+\dfrac{y^2}{\dfrac{41}{9}}=1$
$e=\dfrac{c}{a}=\dfrac{2}{3}$
Đặt $c=2t$, $a=3t$
$\Rightarrow b^2= a^2-b^2=9t^2-4t^2=5t^2$
$(E):\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$
$\Rightarrow \dfrac{x^2}{9t^2}+\dfrac{y^2}{5t^2}=1$
$A\in (E)\Rightarrow \dfrac{1}{9t^2}+\dfrac{2^2}{5t^2}=1$
$\Leftrightarrow t^2=\dfrac{41}{45}$
Vậy $(E):\dfrac{x^2}{\frac{41}{5}}+\dfrac{y^2}{\frac{41}{9}}=1$