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 $A \sqrt{3}$ is equal to
Step 1
The slope of the tangent at any point $(x_{1}, y_{1})$ on the circle is given by $-\frac{x_{1}}{y_{1}}$. Here, the point is $(1, \sqrt{3})$. So, the slope of the tangent is $-\frac{1}{\sqrt{3}}$. Show more…
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