Câu 31:
\(I\left(1;3\right)\in\left(d\right)\Leftrightarrow a+b=3\left(1\right)\)
PT giao Ox: \(y=0\Leftrightarrow x=-\dfrac{b}{a}\Leftrightarrow A\left(-\dfrac{b}{a};0\right)\Leftrightarrow OA=\left|-\dfrac{b}{a}\right|\)
PT giao Oy: \(x=0\Leftrightarrow y=b\Leftrightarrow B\left(0;b\right)\Leftrightarrow OB=\left|b\right|\)
Gọi H là chân đường cao từ O tới đths thì \(OH=\sqrt{5}\)
Áp dụng HTL: \(\dfrac{1}{OA^2}+\dfrac{1}{OB^2}=\dfrac{1}{OH^2}=\dfrac{1}{5}\)
\(\Leftrightarrow\left(\left|-\dfrac{a}{b}\right|\right)^2+\dfrac{1}{\left(\left|b\right|\right)^2}=\dfrac{1}{5}\\ \Leftrightarrow\dfrac{a^2+1}{b^2}=\dfrac{1}{5}\\ \Leftrightarrow5a^2+5=b^2\left(2\right)\)
\(\left(1\right)\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}b=3-a\\5a^2+5=b^2\end{matrix}\right.\Leftrightarrow5a^2+5=\left(3-a\right)^2\\ \Leftrightarrow2a^2+3a-2=0\Leftrightarrow\left[{}\begin{matrix}a=\dfrac{1}{2}\Rightarrow b=\dfrac{5}{2}\\a=-2\Rightarrow b=5\end{matrix}\right.\)
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