The tangent and the normal lines at the point 3 ,   1 to the circle x 2 + y 2 = 4 and the x -axis form…

1. $\frac{1}{3}$
2. $\frac{2}{\sqrt{3}}$
3. $\frac{4}{\sqrt{3}}$
4. $\frac{1}{\sqrt{3}}$

Solution

Equation tangent at $P\left(\sqrt{3},1\right)$
$x\cdot \sqrt{3}+y\cdot 1=4$
x-intercept of this tangent is $\frac{4}{\sqrt{3}}$.
Hence, area of $\Delta =\frac{1}{2}\cdot \frac{4}{\sqrt{3}}\cdot 1=\frac{2}{\sqrt{3}}$

Asked in: JEE Main 2019 (08 Apr Shift 2)

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