Question

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

Answer 1

Since, $\left(sec\right)^{2} \alpha +\left(cosec\right)^{2}⁡\alpha -3=\left(1 + \left(tan\right)^{2} ⁡ \alpha \right)+\left(1 + \left(cot\right)^{2} ⁡ \alpha \right)-3$
$=-1+\left(tan \alpha - cot ⁡ \alpha \right)^{2}+2=1+\left(tan ⁡ \alpha - cot ⁡ \left(\alpha \right)^{2}\right)>0$
Hence, the point $\left(sec \alpha , cos ⁡ e c \alpha \right)$ lies outside the circle $x^{2}+y^{2}-3=0$
Now, the minimum distance
$=\sqrt{\left(sec \alpha - 0\right)^{2} + \left(cosec ⁡ \alpha - 0\right)^{2}}-\sqrt{3}$
$=\sqrt{4 + \left(tan \alpha - cot ⁡ \alpha \right)^{2}}-\sqrt{3}$ (Minimum is attained, when $tan \alpha =cot⁡\alpha $ )
$=2-\sqrt{3}\equiv a-\sqrt{b}$ (Given)
So, $a=2,b=3.$
Hence, $\frac{b}{a}=\frac{3}{2}=1.5$