Question

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 $\alpha\left(0 < \alpha < \frac{\pi}{4}\right)$ with the positive direction of x-axis. The equation of its diagonal not passing through the origin :

• A. $y(\cos \alpha + \sin \, \alpha) + x(\cos \, \alpha - \sin\, \alpha ) = a$
• B. $y(\cos \alpha - \sin \alpha) - x(\sin \alpha - cos \,\alpha) = a$
• C. $y(\cos \alpha + \sin \alpha ) + x(\sin \alpha - \cos \alpha) = a$
• D. $y(\cos \alpha + \sin \alpha) + x(\sin \alpha + \cos \alpha) = a $

Answer 1

The Correct Option is A

Co-ordinates of $A = (a \cos \alpha , a \sin \alpha )$ Equation of OB,

$ y = \tan \left( \frac{\pi}{4} + \alpha \right) x$ $CA \bot^r $ to OB $\therefore $ slope of CA $ = - \cot \left( \frac{\pi}{4} + \alpha \right)$ Equation of CA $y -a \sin\alpha =- \cot\left(\frac{\pi}{4} + \alpha\right) \left(x-a \cos\alpha\right) $ $ \Rightarrow\left(y -a \sin\alpha\right) \left(\tan \left(\frac{\pi}{4} + \alpha\right)\right) = \left(a \cos\alpha-x\right) $ $\Rightarrow \left(y -a \sin\alpha\right) \left(\frac{\tan \frac{\pi}{4}+ \tan\alpha}{1- \tan \frac{\pi}{4} \tan\alpha}\right) \left(a \cos\alpha-x\right)$ $\Rightarrow \left(y -a\sin \alpha\right) \left(1+\tan\alpha\right) = \left(a\cos\alpha-x\right)\left(1 -\tan\alpha\right)$ $ \Rightarrow \left(y -a \sin\alpha\right)\left(\cos\alpha+\sin\alpha\right)=\left(a \cos\alpha - x\right)\left(\cos\alpha - \sin\,\alpha\right)$ $\Rightarrow y \left(\cos+\sin\alpha\right) -a \sin\alpha \cos\alpha -a \sin^{2} \alpha $ $= a \cos^{2} \alpha - a \cos\alpha \sin\alpha - x \left(\cos\alpha - \sin\alpha\right)$ $\Rightarrow y \left(\cos\alpha + \sin\alpha\right) + x\left(\cos\alpha - \sin\alpha\right) = a$ $ y \left(\sin\alpha + \cos\alpha\right) + x \left(\cos\alpha - \sin\alpha\right) = a $.