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

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Question

(b) Co-ordinates of $A = (a\cos \alpha ,\,a\sin \alpha )$
Equation of $OB$, $y = 	an \left( {\frac{\pi }{4} + \alpha } ight)\,x$
$\because$

Vedclass · Accepted Answer

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

Vedclass · Answer

(b) Co-ordinates of $A = (a\cos \alpha ,\,a\sin \alpha )$
Equation of $OB$, $y = 	an \left( {\frac{\pi }{4} + \alpha } ight)\,x$
$\because$  $CA\,\,{ \bot ^r}\,\,{m{to}}\,\,OB$
$	herefore$  Slope of $CA = - \cot \left( {\frac{\pi }{4} + \alpha } ight)$
Equation of CA,$y - a\sin \alpha = - \cot \left( {\frac{\pi }{4} + \alpha } ight)\,(x - a\cos \alpha )$
==> $y(\sin \alpha + \cos \alpha ) + x(\cos \alpha - \sin \alpha ) = a$
$ \Rightarrow $ $y\,(\cos \alpha + \sin \alpha ) - x\,(\sin \alpha - \cos \alpha ) = a$.

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 $x$-axis. The equation of its diagonal not passing through the origin is

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