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)

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

Answer 2

The correct answer is (C) : $4 \sqrt{6}+\frac{28}{3}$
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
1(x+2)2​−16(y−2)2​=1

A=−2∫6​​(2−(x2−4))dx
A=−2∫6​​(6−x2)dx=(6x−3x3​)−26​​
A=(66​−366​​)−(−12+38​)
A=3126​​+328​
A=46​+328​