If the minimum distance of the point (.sec α , cosec â¡ α .) from the circle x2+y2=3 can be expressed as a-√b, where a,b∈ N, then the value of (b/a) is

Question

If the minimum distance of the point (.sec α , cosec â¡ α .) from the circle x2+y2=3 can be expressed as a-√b, where a,b∈ N, then the value of (b/a) is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Since, (sec)2 α +(cosec)2â¡α -3=(1 + (tan)2 â¡ α )+(1 + (cot)2 â¡ α )-3 =-1+(tan α - cot â¡ α )2+2=1+(tan â¡ α - cot â¡ (α )2)>0 Hence, the point (sec α , cos â¡ e c α ) lies outside the circle x2+y2-3=0 Now, the minimum distance =√(sec α - 0)2 + (cosec â¡ α - 0)2-√3 =√4 + (tan α - cot â¡ α )2-√3 (Minimum is attained, when tan α =cotâ¡α ) =2-√3≡ a-√b (Given) So, a=2,b=3. Hence, (b/a)=(3/2)=1.5

Q. If the minimum distance of the point (secα,cosec⁡α) from the circle x2+y2=3 can be expressed as a−b​, where a,b∈N, then the value of ab​ is

Q. If the minimum distance of the point $(sec α, cosec α)$ from the circle $x^{2} + y^{2} = 3$ can be expressed as $a - b,$ where $a, b \in N,$ then the value of $\frac{b}{a}$ is