Area Ques 2
- 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 … .
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Answer:
Correct Answer: 2.( $2 \sqrt{3}$ sq units)
Solution: Equation of tangent at the point $(1, \sqrt{3})$ to the curve
$ x^2+y^2=4 \text { is } x+\sqrt{3} y=4 $
whose $X$-axis intercept $(4,0)$.
Thus, area of $\Delta$ formed by $(0,0)(1, \sqrt{3})$ and $(4,0)$
$=\frac{1}{2}\left|\begin{array}{ccc}0 & 0 & 1 \\ 1 & \sqrt{3} & 1 \\ 4 & 0 & 1\end{array}\right|=\frac{1}{2}|(0-4 \sqrt{3})|=2 \sqrt{3}$ sq units
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