Đáp án:
$\begin{array}{l}
a)AB:y = a.x + b\\
\Rightarrow \left\{ \begin{array}{l}
4 = - 3a + b\\
4 = a + b
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
a = 0\\
b = 4
\end{array} \right.\\
\Rightarrow AB:y = 4\\
b)A,B,O \in \left( C \right)tam\,I\left( {x;y} \right)\\
\Rightarrow I{A^2} = I{B^2} = I{O^2}\\
\Rightarrow {\left( {x + 3} \right)^2} + {\left( {y - 4} \right)^2} = {\left( {x - 1} \right)^2} + {\left( {y - 4} \right)^2}\\
= {x^2} + {y^2}\\
\Rightarrow \left\{ \begin{array}{l}
{x^2} + 6x + 9 + {y^2} - 8y + 16 = {x^2} + {y^2}\\
{x^2} - 2x + 1 + {y^2} - 8y + 16 = {x^2} + {y^2}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
6x - 8y = - 25\\
- 2x - 8y = - 17
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = - 1\\
y = \dfrac{{19}}{8}
\end{array} \right. \Rightarrow I\left( { - 1;\dfrac{{19}}{8}} \right)\\
\Rightarrow \left( C \right):{\left( {x + 1} \right)^2} + {\left( {y - \dfrac{{19}}{8}} \right)^2} = {x^2} + {y^2} = \dfrac{{425}}{{64}}\\
c)I\left( { - 1;\dfrac{{19}}{8}} \right)\\
\Rightarrow \overrightarrow {AI} = \left( {2;\dfrac{{ - 13}}{8}} \right)\\
\Rightarrow PTTT \bot \overrightarrow {AI} :2x - \dfrac{{13}}{8}y + c = 0\\
A \in PTTT \Rightarrow 2.\left( { - 3} \right) - \dfrac{{13}}{8}.4 + c = 0\\
\Rightarrow c = \dfrac{{25}}{2}\\
\Rightarrow PTTT:2x - \dfrac{{13}}{8}y + \dfrac{{25}}{2} = 0\\
d){F_2}\left( {2;0} \right) \Rightarrow c = 2\\
\Rightarrow {a^2} - {b^2} = {c^2} = 4\\
\left( E \right):\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\\
\Rightarrow \dfrac{{{1^2}}}{{{a^2}}} + \dfrac{{{4^2}}}{{{b^2}}} = 1\\
\Rightarrow {b^2} + 16{a^2} = {a^2}{b^2}\\
\Rightarrow \left( {{a^2} - 4} \right) + 16{a^2} = {a^2}\left( {{a^2} - 4} \right)\\
\Rightarrow {a^4} - 21{a^2} + 4 = 0\\
\Rightarrow {a^2} = \dfrac{{21 + 5\sqrt {17} }}{2}\\
\Rightarrow {b^2} = {a^2} - 4 = \dfrac{{13 + 5\sqrt {17} }}{2}\\
\Rightarrow \left( E \right):\dfrac{{2{x^2}}}{{21 + 5\sqrt {17} }} + \dfrac{{2{y^2}}}{{13 + 5\sqrt {17} }} = 1
\end{array}$
