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

Answer: (A)

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

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