Question

The foot of the perpendicular drawn from $(0,0)$ to the line $x\cos \alpha +y\sin \alpha -p=0$ is

• A. $(p\sin \alpha ,p\cos \alpha )$
• B. $(p\sec \alpha ,p\cos \alpha )$
• C. $(p\cos \alpha ,p\sin \alpha )$
• D. $(p,p)$

Answer 1

The correct option is C $(p\cos \alpha ,p\sin \alpha )$

$\therefore x\cos \alpha +y\sin \alpha -p=0$

From the formula for foot of perpendicular,

$\frac{x_2-0}{\cos \alpha }=\frac{y_2-0}{\sin \alpha }=-\frac{(0\times \cos \alpha +0\times \sin \alpha -p)}{({\cos }^2\alpha +{\sin }^2\alpha )}=p$

$x_2=\frac{+p\cos \alpha }{1}=p\cos \alpha$

$y_2=p\sin \alpha$