The minimum distance between the curves y=tan x,∀ x∈ (- (π /2) , (π /2)) and (x - 2 - (π /4))2+y2=1 is

Question

The minimum distance between the curves y=tan x,∀ x∈ (- (π /2) , (π /2)) and (x - 2 - (π /4))2+y2=1 is (A) √2-1 (B) √5-1 (C) √5+1 (D) 2

Tardigrade Admin · Accepted Answer

The minimum distance between two curves lies along their common normal.
Let,

P (h, t anh)

lies on

y = t an x

Then, the equation of normal at

P

is

y - t anh = - \frac{1}{se c ^{2} h} (x - h)

this passes through the centre of the circle, hence,

t anh se c^{2} h = 2 + \frac{π}{4} - h

\Rightarrow h = \frac{π}{4}

Minimum distance = distance between

(\frac{π}{4}, 1)

and

(2 + \frac{π}{4}, 0)

- 1

= 5 - 1

Q. The minimum distance between the curves $y = t an x, \forall x \in (- \frac{π}{2}, \frac{π}{2})$ and $(x - 2 - \frac{π}{4})^{2} + y^{2} = 1$ is