question_answer

The tangent and the normal lines at the point \[\left( \sqrt{3},1 \right)\] to the circle \[{{x}^{2}}+{{y}^{2}}=4\] and the x-axis form a triangle. The area of this triangle (in square units) is:             [JEE Main 8-4-2019 Afternoon]

A) \[\frac{1}{3}\]                        

B) \[\frac{4}{\sqrt{3}}\]

C) \[\frac{1}{\sqrt{3}}\]             

D)   \[\frac{2}{\sqrt{3}}\]

Correct Answer: D

Solution :

Given \[{{x}^{2}}+{{y}^{2}}=4\] equation of tangent \[\Rightarrow \]\[\sqrt{3}x+y=4\]                                           ...(1) Equation of normal \[x-\sqrt{3}y=0\] ...(2) Coordinate of \[T\left( \frac{4}{\sqrt{3}},0 \right)\] \[\therefore \]Area of triangle \[=\frac{2}{\sqrt{3}}\]