Question

Foot of the perpendicular from $(-2,\frac{\pi }{4})$ on $r[\sqrt{2}\cos \theta +(3+\sqrt{2}\left)\sin \theta \right]=9+3\sqrt{2}$ is:

• A. $(3,\frac{\pi }{2})$
• B. $(3,\frac{\pi }{4})$
• C. $(\sqrt{2},\frac{\pi }{2})$
• D. $(\sqrt{2},\frac{\pi }{4})$

Answer 1

The correct option is A $(3,\frac{\pi }{2})$

The Cartesian form of $(-2,\frac{\pi }{4})$ is $(-2\cos \frac{\pi }{4},-2\sin \frac{\pi }{4})\equiv (-\sqrt{2},-\sqrt{2})$

The line equation is $\sqrt{2}r\cos \theta +(3+\sqrt{2})r\sin \theta =9+3\sqrt{2}$

$\Rightarrow \sqrt{2}x+(3+\sqrt{2})y=9+3\sqrt{2}$

Let the foot of perpendicular be $(h,k)$

We have $\sqrt{2}h+(3+\sqrt{2})k=9+3\sqrt{2}$ and $\frac{k+\sqrt{2}}{h+\sqrt{2}}\times (-\frac{\sqrt{2}}{3+\sqrt{2}})=-1$

$\Rightarrow \sqrt{2}h+(3+\sqrt{2})k=9+3\sqrt{2}$ and $(3+\sqrt{2})h+3\sqrt{2}=k\sqrt{2}$

By solving above equations, we get $h=0,k=3$

Therefore, polar form of feet of perpendicular is $(3,\frac{\pi }{2})$.