A mirror in the first quadrant is in the shap | vedclass

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(C) The equation of the hyperbola is $xy=1$,which can be written as $y = \frac{1}{x}$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = -\fr

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$\frac{15}{8}$

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(C) The equation of the hyperbola is $xy=1$,which can be written as $y = \frac{1}{x}$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = -\frac{1}{x^2}$.The slope of the tangent at the point $(2, 1/2)$ is $\left(\frac{dy}{dx}ight)_{(2, 1/2)} = -\frac{1}{2^2} = -\frac{1}{4}$.The slope of the normal at $(2, 1/2)$ is $n = -\frac{1}{(-1/4)} = 4$.Let $m$ be the slope of the incident beam of light. The reflected ray is parallel to the $Y$-axis,so its slope is $\infty$.According to the law of reflection,the angle between the incident ray and the normal is equal to the angle between the reflected ray and the normal. Using the formula for the angle between two lines with slopes $m_1$ and $m_2$,$	an 	heta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}ight|$,we have:$\left|\frac{4 - m}{1 + 4m}ight| = \left|\frac{\infty - 4}{1 + 4(\infty)}ight| = \left|\frac{1}{4}ight|$.Solving $\frac{4 - m}{1 + 4m} = \frac{1}{4}$:$16 - 4m = 1 + 4m$$8m = 15 \Rightarrow m = \frac{15}{8}$.Solving $\frac{4 - m}{1 + 4m} = -\frac{1}{4}$:$16 - 4m = -1 - 4m$$16 = -1$,which is impossible.Thus,the slope of the incident beam is $\frac{15}{8}$.

$A$ mirror in the first quadrant is in the shape of a hyperbola whose equation is $xy=1$. $A$ light source in the second quadrant emits a beam of light that hits the mirror at the point $(2, 1/2)$. If the reflected ray is parallel to the $Y$-axis,the slope of the incident beam is

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