A square of side a lies above the x-axis and | vedclass

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(B) Let the vertices of the square be $O(0,0)$,$A(a \cos \alpha, a \sin \alpha)$,$B$,and $C$. Since the side $OA$ makes an angle $\alpha$ with the $x$

Vedclass · Accepted Answer

$y(\cos \alpha + \sin \alpha) - x(\sin \alpha - \cos \alpha) = a$

Vedclass · Answer

(B) Let the vertices of the square be $O(0,0)$,$A(a \cos \alpha, a \sin \alpha)$,$B$,and $C$. Since the side $OA$ makes an angle $\alpha$ with the $x$-axis,its coordinates are $A(a \cos \alpha, a \sin \alpha)$. The diagonal $OB$ makes an angle of $\alpha + \frac{\pi}{4}$ with the $x$-axis. The slope of $OB$ is $	an(\alpha + \frac{\pi}{4})$. The diagonal $AC$ is perpendicular to $OB$. The slope of $AC$ is $-\cot(\alpha + \frac{\pi}{4}) = -\frac{1}{	an(\alpha + \frac{\pi}{4})} = -\frac{1 - 	an \alpha 	an(\frac{\pi}{4})}{	an \alpha + 	an(\frac{\pi}{4})} = -\frac{1 - 	an \alpha}{1 + 	an \alpha} = \frac{	an \alpha - 1}{	an \alpha + 1} = \frac{\sin \alpha - \cos \alpha}{\sin \alpha + \cos \alpha}$. The equation of the line $AC$ passing through $A(a \cos \alpha, a \sin \alpha)$ is: $y - a \sin \alpha = \frac{\sin \alpha - \cos \alpha}{\sin \alpha + \cos \alpha} (x - a \cos \alpha)$ $y(\sin \alpha + \cos \alpha) - a \sin \alpha(\sin \alpha + \cos \alpha) = x(\sin \alpha - \cos \alpha) - a \cos \alpha(\sin \alpha - \cos \alpha)$ $y(\sin \alpha + \cos \alpha) - x(\sin \alpha - \cos \alpha) = a \sin^2 \alpha + a \sin \alpha \cos \alpha - a \sin \alpha \cos \alpha + a \cos^2 \alpha$ $y(\sin \alpha + \cos \alpha) - x(\sin \alpha - \cos \alpha) = a(\sin^2 \alpha + \cos^2 \alpha) = a$. Thus,the equation is $y(\cos \alpha + \sin \alpha) - x(\sin \alpha - \cos \alpha) = a$.

$A$ square of side $a$ lies above the $x$-axis and has one vertex at the origin. The side passing through the origin makes an angle $\alpha, (0 < \alpha < \frac{\pi}{4})$ with the positive direction of the $x$-axis. The equation of its diagonal not passing through the origin is

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