If the minimum distance of the point (sec α , cosec α ) from the circle x2+y2=3 can be expressed as a-√b, where a,b∈ N, then the value of a3+b2 is

Question

If the minimum distance of the point (sec α , cosec α ) from the circle x2+y2=3 can be expressed as a-√b, where a,b∈ N, then the value of a3+b2 is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Since, sec 2 α+ operatornamecosec2 α-3=(1+ tan 2 α)+(1+ cot 2 α)-3 =-1+( tan α- cot α)2+2=1+( tan α- cot α2)>0 Hence, the point ( sec α, operatornamecosec α) lies outside the circle x2+y2-3=0 Now, the minimum distance beginarrayl =√( sec α-0)2+( operatornamecosec α-0)2-√3 .=√4+( tan α- cot α)2-√3 text (Minimum is attained, when tan α= cot α) =2-√3 ≡ a-√b text (Given) text So, a=2, b=3 text Hence, a3+b2=17 endarray

Q. If the minimum distance of the point (secα,cosecα) from the circle x2+y2=3 can be expressed as a−b​, where a,b∈N, then the value of a3+b2 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 $a^{3} + b^{2}$ is