Let y = p (x) be the parabola passing through the points (– 1, 0),...

Let, the parabola be $y = a{x}^{2} + bx + c$,

Now given parabola is passing through \left(- 1, 0\right), \left(0, 1\right) \text{$and$} \left(1, 0\right)$</p><p>Now putting the value in the above equation of parabola we get,</p><p>$a - b + c = 0 . . . . . . \left(1\right), $c = 1 . . . . \left(2\right)$ and a + b + c = 0 . . . .\left(3\right)

Now solving above three equations we get,

Parabola as $y = 1 - {x}^{2}$

Now area of region between $y = 1 - {x}^{2} and {\left(x + 1\right)}$^{2} + {\left(y - 1\right)}^{2} \le 1is given by,</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/5f6a2e8d-9d16-4fee-bdda-5b832bc2d93b.png" style="width: 350px; height: 251px;" /></p><p>Now from above diagram, required area will be,

A = {∫}_{- 1}^{0}\left(- {x}^{2} + \sqrt{1 - {\left(x + 1\right)}^{2}}\right)dx$</p><p>$⇒ A = \dfrac{\pi}{4} - \dfrac{1}{3}$</p><p>Hence,$12 ( \pi - 4A ) = 16