In Fig, AOBA is the part of the ellipse $9x^2 + y^2 = 36$ in the first quadrant such that $OA = 2$ and $OB = 6$. Find the area between the arc AB and the chord AB.

$3\pi - 6$

The correct answer is Option (1) → $3\pi - 6$

Given equation of the ellipse $9x^2 + y^2 = 36$ can be expressed as $\frac{x^2}{4} + \frac{y^2}{36} = 1$ or $\frac{x^2}{2^2} + \frac{y^2}{6^2} = 1$ and hence, its shape is as given in Fig.

Accordingly, the equation of the chord AB is

$y - 0 = \frac{6 - 0}{0 - 2}(x - 2)$

$\text{or } \quad y = -3(x - 2)$

$\text{or } \quad y = -3x + 6$

Area of the shaded region as shown in the Fig.

$= 3 \int\limits_0^2 \sqrt{4 - x^2} \, dx - \int\limits_0^2 (6 - 3x) \, dx$

$= 3 \left[ \frac{x}{2} \sqrt{4 - x^2} + \frac{4}{2} \sin^{-1} \frac{x}{2} \right]_0^2 - \left[ 6x - \frac{3x^2}{2} \right]_0^2$

$= 3 \left[ \frac{2}{2} \times 0 + 2 \sin^{-1}(1) \right] - \left[ 12 - \frac{12}{2} \right] = 3 \times 2 \times \frac{\pi}{2} - 6 = 3\pi – 6$