JEE Main & Advanced
JEE Main Paper (Held on 08-4-2019 Afternoon)
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 \[=\frac{2}{\sqrt{3}}\]