Đáp án: $\left( E \right):\dfrac{{{x^2}}}{{18,57}} + \dfrac{{{y^2}}}{{12,57}} = 1$
Giải thích các bước giải:
Gọi pt chính tắc của (E) là
$\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$
$\begin{array}{l}
M\left( {3;\dfrac{{16}}{5}} \right) \in \left( E \right)\\
\Leftrightarrow \dfrac{{{3^2}}}{{{a^2}}} + \dfrac{{{{16}^2}}}{{{5^2}.{b^2}}} = 1\\
\Leftrightarrow \dfrac{9}{{{a^2}}} + \dfrac{{256}}{{25{b^2}}} = 1\\
Do:c = 6\\
\Leftrightarrow {a^2} - {b^2} = {c^2} = 36\\
\Leftrightarrow {a^2} = {b^2} + 36\\
\Leftrightarrow \dfrac{9}{{{b^2} + 36}} + \dfrac{{256}}{{25{b^2}}} = 1\\
\Leftrightarrow \dfrac{{9.25{b^2} + 256.\left( {{b^2} + 36} \right)}}{{25{b^2}\left( {{b^2} + 36} \right)}} = 1\\
Dat:{b^2} = x\\
\Leftrightarrow \dfrac{{225x + 256x + 256.36}}{{25.x\left( {x + 36} \right)}} = 1\\
\Leftrightarrow 481x + 9216 = 25{x^2} + 900x\\
\Leftrightarrow 25{x^2} + 419x - 9216 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 12,57\left( {tm} \right)\\
x = - 29,33\left( {ktm} \right)
\end{array} \right.\\
\Leftrightarrow {b^2} = 12,57\\
\Leftrightarrow {a^2} = {b^2} + 6 = 18,57\\
\Leftrightarrow \left( E \right):\dfrac{{{x^2}}}{{18,57}} + \dfrac{{{y^2}}}{{12,57}} = 1
\end{array}$
