Question

Let $T$ and $C$ respectively be the transverse and conjugate axes of the hyperbola $16 x^2-y^2+64 x+4 y$ $+44=0$. Then the area of the region above the parabola $x^2=y+4$, below the transverse axis $T$ and on the right of the coujugate axis $C$ is:

• A. $4 \sqrt{6}+\frac{44}{3}$
• B. $4 \sqrt{6}-\frac{28}{3}$
• C. $4 \sqrt{6}+\frac{28}{3}$
• D. $4 \sqrt{6}-\frac{44}{3}$

Answer 1

Answer: (C)

$ 16\left(x^2+4 x\right)-\left(y^2-4 y\right)+44=0 $
$ 16(x+2)^2-64-(y-2)^2+4+44=0$
$ 16(x+2)^2-(y-2)^2=16 $
$\frac{(x+2)^2}{1}-\frac{(y-2)^2}{16}=1$

$ A=\int\limits_{-2}^{\sqrt{6}}\left(2-\left(x^2-4\right)\right) d x$
$ A=\int\limits_{-2}^{\sqrt{6}}\left(6-x^2\right) d x=\left(6 x-\frac{x^3}{3}\right)_{-2}^{\sqrt{6}}$
$ A=\left(6 \sqrt{6}-\frac{6 \sqrt{6}}{3}\right)-\left(-12+\frac{8}{3}\right) $
$A=\frac{12 \sqrt{6}}{3}+\frac{28}{3} $
$ A=4 \sqrt{6}+\frac{28}{3}$