Let T and C respectively be the transverse and conjugate axes of the hyperbola 16x2−y2+64x+4y+44=0. Then the area of the region above the parabola x2=y+4, below the transverse axis T, and on the right of the conjugate axis C is:
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Official solution
Completing the square: 16(x2+4x)−(y2−4y)+44=0⇒16(x+2)2−64−(y−2)2+4+44=0⇒16(x+2)2−(y−2)2=16, i.e.
\[\frac{(x+2)^2}{1}-\frac{(y-2)^2}{16}=1.\]
Centre (−2,2); transverse axis is the horizontal line y=2, conjugate axis is the vertical line x=−2.
The parabola x2=y+4, i.e. y=x2−4, meets y=2 where x2−4=2⇒x=±6. The required region (above the parabola, below y=2, right of x=−2) runs from x=−2 to x=6:
\[A=\int_{-2}^{\sqrt{6}}\left[2-(x^2-4)\right]dx=\int_{-2}^{\sqrt{6}}(6-x^2)\,dx=\left[6x-\frac{x^3}{3}\right]_{-2}^{\sqrt{6}}\]
\[=\left(6\sqrt{6}-2\sqrt{6}\right)-\left(-12+\frac83\right)=4\sqrt{6}+\frac{28}{3}.\]