Normals
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Find the equation of the normal to the curve \(y=x+\dfrac{1}{x}\) for \(x=2\)
\(\begin{aligned}
&y=x+x^{-1} \\
&\dfrac{d y}{d x}=1-x^{-2}=1-\dfrac{1}{x^2}
\end{aligned}\)
At \(x=2 \quad \dfrac{d y}{d x}=1-\dfrac{1}{4}=\dfrac{3}{4}\)
At \(x=2 \quad y=1+\dfrac{1}{2}=\dfrac{3}{2}\)
gradient of normal \(=-\dfrac{4}{3}\)
\(\begin{aligned}
&y-y_1=m\left(x-x_1\right) \\
&y-\dfrac{3}{2}=-\dfrac{4}{3}(x-2) \\
&6\left(y-\dfrac{3}{2}\right)=6\left(-\dfrac{4}{3}(x-2)\right) \\
&6 y-9=-8(x-2) \\
&6 y-9=-8 x+16 \\
&\therefore 8 x+6 y-25=0 \\
&\text { is the equation of the normal }
\end{aligned}\)
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