Question

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

Answer 1

Since, $\sec ^{2} \alpha+\operatorname{cosec}^{2} \alpha-3=\left(1+\tan ^{2} \alpha\right)+\left(1+\cot ^{2} \alpha\right)-3$
$=-1+(\tan \alpha-\cot \alpha)^{2}+2=1+\left(\tan \alpha-\cot \alpha^{2}\right)>0$
Hence, the point $(\sec \alpha, \operatorname{cosec} \alpha)$ lies outside the circle $x^{2}+y^{2}-3=0$
Now, the minimum distance
$ \begin{array}{l} =\sqrt{(\sec \alpha-0)^{2}+(\operatorname{cosec} \alpha-0)^{2}}-\sqrt{3} \\ \left.=\sqrt{4+(\tan \alpha-\cot \alpha)^{2}}-\sqrt{3} \text { (Minimum is attained, when } \tan \alpha=\cot \alpha\right) \\ =2-\sqrt{3} \equiv a-\sqrt{b} \text { (Given) } \\ \text { So, } a=2, b=3 \\ \text { Hence, } a^{3}+b^{2}=17 \end{array} $