Đáp án:
\(\left[ \begin{array}{l}
\left( P \right):\,\,\,y = \frac{1}{6}{x^2} + \frac{{11}}{6}x - 7\\
\left( P \right):\,\,\,y = - \frac{1}{6}{x^2} + \frac{{25}}{6}x - 11
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
E\left( {4;\,\,3} \right) \in \left( P \right) \Rightarrow 3 = 16a + 4b + c\\
F\left( {3;\,\,0} \right) \in \left( P \right) \Rightarrow 0 = 9a + 3b + c\\
Oy \cap \left( P \right) = \left\{ P \right\} \Rightarrow P\left( {0;\,\,c} \right)\\
Ta\,\,co:\,\,\,\overrightarrow {EF} = \left( { - 1;\, - 3} \right) \Rightarrow EF = \sqrt {10} .\\
Pt\,\,duong\,\,thang\,\,\,EF:\,\,\frac{{x - 3}}{{4 - 3}} = \frac{y}{3}\\
\Leftrightarrow 3x - 9 = y \Leftrightarrow EF:\,\,\,3x - y - 9 = 0\\
\Rightarrow {S_{EFP}} = \frac{1}{2}d\left( {P;\,\,EF} \right).EF = 1\\
\Leftrightarrow \frac{{\left| { - c - 9} \right|}}{{\sqrt {{3^2} + 1} }}.\sqrt {10} = 2 \Leftrightarrow \left| {c + 9} \right| = 2 \Leftrightarrow \left[ \begin{array}{l}
c + 9 = 2\\
x + 9 = - 2
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
c = - 7\\
c = - 11
\end{array} \right..\\
+ )\,\,Voi\,\,c = - 8\\
\Rightarrow \left\{ \begin{array}{l}
9a + 3b - 7 = 0\\
16a + 4b - 7 = 3
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a = \frac{1}{6}\\
b = \frac{{11}}{6}
\end{array} \right. \Rightarrow \left( P \right):\,\,\,y = \frac{1}{6}{x^2} + \frac{{11}}{6}x - 7\\
+ )\,\,\,Voi\,\,x = - 11\\
\Rightarrow \left\{ \begin{array}{l}
9a + 3b - 11 = 0\\
16a + 4b - 11 = 3
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a = - \frac{1}{6}\\
b = \frac{{25}}{6}
\end{array} \right. \Rightarrow \left( P \right):\,\,\,y = - \frac{1}{6}{x^2} + \frac{{25}}{6}x - 11
\end{array}\)
