A(1, -3) and B(4, 3) are two points on the cu | vedclass

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(B) The slope of the chord $AB$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - (-3)}{4 - 1} = \frac{6}{3} = 2$. Since the tangent is paralle

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$(2, 0), (-2, 0)$

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(B) The slope of the chord $AB$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - (-3)}{4 - 1} = \frac{6}{3} = 2$. Since the tangent is parallel to the chord $AB$,the slope of the tangent must also be $2$. Given the curve $y = x - \frac{4}{x}$,we find the derivative $\frac{dy}{dx} = 1 + \frac{4}{x^2}$. Setting the derivative equal to the slope of the chord: $1 + \frac{4}{x^2} = 2$. This simplifies to $\frac{4}{x^2} = 1$,which implies $x^2 = 4$,so $x = \pm 2$. For $x = 2$,$y = 2 - \frac{4}{2} = 0$. For $x = -2$,$y = -2 - \frac{4}{-2} = -2 + 2 = 0$. Thus,the required points are $(2, 0)$ and $(-2, 0)$.

$A(1, -3)$ and $B(4, 3)$ 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 $AB$,are

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