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$\mathrm{A}(1,-3), \mathrm{B}(4,3)$ are two points on the curve $y=x-\frac{4}{x}$. The points on the curve, the tangents at which are parallel to the chord $\mathrm{AB}$, are
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$(2,0),(-2,0)$
Slope of tangent $=$ slope of $\mathrm{AB}=\frac{3+3}{4-1}=\frac{6}{3}=2$
$y=x-\frac{4}{x}$
$\therefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=1+\frac{4}{x^2}$
$\begin{aligned} & \Rightarrow 2=1+\frac{4}{x^2} \\ & \Rightarrow x^2=4 \\ & \Rightarrow x= \pm 2\end{aligned}$
When $x=2, y=2-\frac{4}{2}=0$
When $x=-2, y=-2+\frac{4}{2}=0$
$\therefore \quad$ The required points are $(2,0)$ and $(-2,0)$.
$y=x-\frac{4}{x}$
$\therefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=1+\frac{4}{x^2}$
$\begin{aligned} & \Rightarrow 2=1+\frac{4}{x^2} \\ & \Rightarrow x^2=4 \\ & \Rightarrow x= \pm 2\end{aligned}$
When $x=2, y=2-\frac{4}{2}=0$
When $x=-2, y=-2+\frac{4}{2}=0$
$\therefore \quad$ The required points are $(2,0)$ and $(-2,0)$.
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