Question

If $A$ is the area of the triangle formed by positive $x$-axis and the normal and the tangent to the circle $x^2+y^2=4$ at $(1,\sqrt{3})$ then $\frac{A}{\sqrt{3}}$ is equal to

Answer 1

Equation of tangent to the given circle at $(1,\sqrt{3})$ is,
$T:x+\sqrt{3}y=4$
and normal at the same point is,
$(y-0)=\sqrt{3}(x-0)\Rightarrow y=\sqrt{3}x$,
since $(0,0)$ is the centre of the circle.
Thus required area $=$ area of $\Delta OAB=\frac{1}{2}\times 4\times \sqrt{3}=2\sqrt{3}$