Question

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 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)

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

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