Let T and C respectively, be the transverse and conjugate axes of...

Given,

Equation of hyperbola,

$16\left({x}^{2} + 4x\right) - \left({y}^{2} - 4y\right) + 44 = 0$

$⇒ 16 ( x + 2{)}^{2} - 64 - ( y - 2{)}^{2} + 4 + 44 = 0$

$⇒ 16{\left(x + 2\right)}^{2} - {\left(y - 2\right)}^{2} = 16$

$⇒ \dfrac{{\left(x + 2\right)}^{2}}{1} - \dfrac{{\left(y - 2\right)}^{2}}{16} = 1$

Hence, equation of conjugate axis will be $x = - 2$ and equation of transverse axis is given by $y = 2$,

Now plotting the diagram of parabola ${x}^{2} = y + 4$ and $x = - 2 \text{ and } y = 2$ we get,

Now from above diagram, the area of the bounded region is given by,

$A = {∫}_{- 2}^{\sqrt{6}}\left(2 - \left({x}^{2} - 4\right)\right)dx$

$⇒ A = {∫}_{- 2}^{\sqrt{6}}\left(6 - {x}^{2}\right)dx = {\left(6x - \dfrac{{x}^{3}}{3}\right)}_{- 2}^{\sqrt{6}}$

$⇒ A = \left(6\sqrt{6} - \dfrac{6\sqrt{6}}{3}\right) - \left(- 12 + \dfrac{8}{3}\right)$

$⇒ A = \dfrac{12\sqrt{6}}{3} + \dfrac{28}{3}$

$⇒ A = 4\sqrt{6} + \dfrac{28}{3}$