Question

The minimum distance between the curves $y=tan x,\forall x\in \left(- \frac{\pi }{2} , \frac{\pi }{2}\right)$ and $\left(x - 2 - \frac{\pi }{4}\right)^{2}+y^{2}=1$ is

• A. $\sqrt{2}-1$
• B. $\sqrt{5}-1$
• C. $\sqrt{5}+1$
• D. $2$

Answer 1

Answer: (B)

The minimum distance between two curves lies along their common normal.
Let, $P\left(h , t a n h\right)$ lies on $y=tan x$
Then, the equation of normal at $P$
is $y-tan h=-\frac{1}{s e c^{2} h}\left(x - h\right)$
this passes through the centre of the circle, hence,
$tan hsec^{2} ⁡ h=2+\frac{\pi }{4}-h$
$\Rightarrow h=\frac{\pi }{4}$
Minimum distance = distance between $\left(\frac{\pi }{4} , 1\right)$ and $\left(2 + \frac{\pi }{4} , 0\right)$ $-1$
$=\sqrt{5}-1$