Question

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

• A. $2\sqrt{3}$
• B. $2$
• C. $\sqrt{3}$
• D. $6$

Answer 1

The correct option is B $2$

We have,
$x^2+y^2=4$
$\Rightarrow 2x+2y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac{x}{y}$
${\Rightarrow \frac{dy}{dx}|}_{(1,\sqrt{3})}=-\frac{1}{\sqrt{3}}$
Therefore, the equation of the tangent at $(1,\sqrt{3})$ is
$y-\sqrt{3}=-\frac{1}{\sqrt{3}}(x-1)$
and the point of intersection of this tangent with th $x$ -axis is $(4,0)$.
The equation of the normal at $(1,\sqrt{3})$
$y-\sqrt{3}=\sqrt{3}(x-1)$,
and the point of intersection of the normal with the $x$ -axis is $(0,0)$.
Hence the required area is:
$A=\frac{1}{2}\cdot 4\cdot \sqrt{3}=2\sqrt{3}$
$\therefore \frac{A}{\sqrt{3}}=2$