Problem

Problem

If tangent and normal to the curve y=2sin⁡x+sin⁡2xy = 2 \sin x + \sin 2x are drawn at P(x=π3)P \left( x = \frac{\pi}{3} \right), then area of the quadrilateral formed by the tangent, the normal at PP and the coordinate axes is a. π3\frac{\pi}{3} b. 3π3\pi c. π32\frac{\pi\sqrt{3}}{2} d. None of these
Intuitive explanation

The key is finding where the tangent and normal lines intersect the coordinate axes, then using those four intersection points as vertices of your quadrilateral. Start by computing the derivative to get the slope at x=π3x = \frac{\pi}{3}, then write the equations of both lines. The tangent and normal will each cross both axes at different points—those four crossings, along with the origin, form your quadrilateral. Finally, use the standard area formula (or break it into triangles) to find the enclosed region.

Intuitive explanation

The key is finding where the tangent and normal lines intersect the coordinate axes, then using those four intersection points as vertices of your quadrilateral. Start by computing the derivative to get the slope at x=π3x = \frac{\pi}{3}, then write the equations of both lines. The tangent and normal will each cross both axes at different points—those four crossings, along with the origin, form your quadrilateral. Finally, use the standard area formula (or break it into triangles) to find the enclosed region.

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