The cross-section of a reflecting surface is | vedclass

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(C) The equation of the reflecting surface is $x^{2}+y^{2}=R^{2}$.Differentiating with respect to $x$,we get $2x + 2y \frac{dy}{dx} = 0$,which implies

Vedclass · Accepted Answer

$\left(-\frac{R}{\sqrt{2}}, \frac{R}{\sqrt{2}}\right)$

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(C) The equation of the reflecting surface is $x^{2}+y^{2}=R^{2}$.Differentiating with respect to $x$,we get $2x + 2y \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx} = -\frac{x}{y}$.This $\frac{dy}{dx}$ represents the slope of the tangent at point $M$. The normal at point $M$ makes an angle of $45^{\circ}$ with the $x$-axis because the incident ray is horizontal and the reflected ray is vertical,making the angle of incidence equal to the angle of reflection ($22.5^{\circ}$ each relative to the normal).Alternatively,for the ray to turn $90^{\circ}$ (from $+x$ to $+y$),the normal at $M$ must make an angle of $45^{\circ}$ with the $x$-axis.The slope of the normal is $	an(45^{\circ}) = 1$. Since the normal is perpendicular to the tangent,the slope of the tangent is $-1$.Thus,$-\frac{x}{y} = -1$,which means $x = y$.Substituting $x = y$ into the equation $x^{2}+y^{2}=R^{2}$,we get $2x^{2} = R^{2}$,so $x = \pm \frac{R}{\sqrt{2}}$.Based on the geometry of the reflection (ray coming from left,reflecting to $+y$),point $M$ must be in the second quadrant,where $x  0$.Therefore,the coordinates are $M = \left(-\frac{R}{\sqrt{2}}, \frac{R}{\sqrt{2}}ight)$.

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