Question

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

Answer 1

The tangent at $P(1,\sqrt{3})$ is $x+y\sqrt{3}=4$ and normal is $\sqrt{3}x-y=0$. They form a $△OPA$ with the x-axis.
Clearly $OA=4$ and $PN=\sqrt{3}$ so that the area of $△OPA=\frac{1}{2}OA.PN=\frac{1}{2}.4.\sqrt{3}=2\sqrt{3}$.