Let T and C respectively, be the transverse and conjugate axes of the hyperbola 16 x 2 - y 2 + 64 x + 4 y +…

Given,

Equation of hyperbola,

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

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

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

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

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 $x=-2 \& y=2$ we get,

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

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

$\Rightarrow A={\int }_{-2}^{\sqrt{6}}\left(6-x^2\right)dx={\left(6x-\frac{x^3}{3}\right)}_{-2}^{\sqrt{6}}$

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

$\Rightarrow A=\frac{12\sqrt{6}}{3}+\frac{28}{3}$

$\Rightarrow A=4\sqrt{6}+\frac{28}{3}$