Let T and C respectively, be the transverse and conjugate axes of the hyperbola 16x^2-y^2+64x+4y+44=0. Then the area of the region above the parabola…

Given,

Equation of hyperbola,

$$16({x}^{2}+4x)-({y}^{2}-4y)+44=0$$16(x2+4x)−(y2−4y)+44=0

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

$$\Rightarrow 16{(x+2)}^{2}-{(y-2)}^{2}=16$$⇒16(x+2)2−(y−2)2=16

$$\Rightarrow \frac{{(x+2)}^{2}}{1}-\frac{{(y-2)}^{2}}{16}=1$$⇒1(x+2)2​−16(y−2)2​=1

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

Now plotting the diagram of parabola ${x}^{2}=y+4$x2=y+4 and x=-2&y=2 we get,

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

$$A={\int }_{-2}^{\sqrt{6}}(2-({x}^{2}-4))dx$$A=∫−26​​(2−(x2−4))dx

$$\Rightarrow A={\int }_{-2}^{\sqrt{6}}(6-{x}^{2})dx={(6x-\frac{{x}^{3}}{3})}_{-2}^{\sqrt{6}}$$⇒A=∫−26​​(6−x2)dx=(6x−3x3​)−26​​

$$\Rightarrow A=(6\sqrt{6}-\frac{6\sqrt{6}}{3})-(-12+\frac{8}{3})$$⇒A=(66​−366​​)−(−12+38​)

$$\Rightarrow A=\frac{12\sqrt{6}}{3}+\frac{28}{3}$$⇒A=3126​​+328​

$$\Rightarrow A=4\sqrt{6}+\frac{28}{3}$$⇒A=46​+328​