viết phương trình đường thẳng : y=ax^2+bx+c biết: P đi qua E (4;3), cắt Ox tại F (3;0), cắt Oy tại P sao cho tam giác EFP có diện tích bằng 1

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