Let T and C respectively be the transverse an | vedclass

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(B) First,rewrite the hyperbola equation in standard form:$16(x^2 + 4x) - (y^2 - 4y) + 44 = 0$$16(x+2)^2 - 64 - (y-2)^2 + 4 + 44 = 0$$16(x+2)^2 - (y-2

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$4 \sqrt{6} + \frac{28}{3}$

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(B) First,rewrite the hyperbola equation in standard form:$16(x^2 + 4x) - (y^2 - 4y) + 44 = 0$$16(x+2)^2 - 64 - (y-2)^2 + 4 + 44 = 0$$16(x+2)^2 - (y-2)^2 = 16$$\frac{(x+2)^2}{1} - \frac{(y-2)^2}{16} = 1$The transverse axis $T$ is the line $y = 2$ and the conjugate axis $C$ is the line $x = -2$.The parabola is $y = x^2 - 4$.The region is bounded by $y = 2$ (above),$y = x^2 - 4$ (below),and $x = -2$ (left).To find the intersection of $y = 2$ and $y = x^2 - 4$,set $x^2 - 4 = 2$,which gives $x^2 = 6$,so $x = \sqrt{6}$ (since we are on the right of $x = -2$).The area $A$ is given by:$A = \int_{-2}^{\sqrt{6}} (2 - (x^2 - 4)) dx$$A = \int_{-2}^{\sqrt{6}} (6 - x^2) dx$$A = [6x - \frac{x^3}{3}]_{-2}^{\sqrt{6}}$$A = (6\sqrt{6} - \frac{6\sqrt{6}}{3}) - (-12 - \frac{-8}{3})$$A = (6\sqrt{6} - 2\sqrt{6}) - (-12 + \frac{8}{3})$$A = 4\sqrt{6} - (-\frac{28}{3}) = 4\sqrt{6} + \frac{28}{3}$

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 $x^2 = y + 4$,below the transverse axis $T$ and on the right of the conjugate axis $C$ is:

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