Hyperbola Question 3
Question 3 - 25 January - Shift 2
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 conjugate axis $C$ is:
(1) $4 \sqrt{6}+\frac{44}{3}$
(2) $4 \sqrt{6}+\frac{28}{3}$
(3) $4 \sqrt{6}-\frac{44}{3}$
(4) $4 \sqrt{6}-\frac{28}{3}$
Show Answer
Answer: (2)
Solution:
Formula: Conjugate axis of hyperbola, Area under curves Formula
$16(x^{2}+4 x)-(y^{2}-4 y)+44=0$
$16(x+2)^{2}-64-(y-2)^{2}+4+44=0$
$16(x+2)^{2}-(y-2)^{2}=16$
$\frac{(x+2)^{2}}{1}-\frac{(y-2)^{2}}{16}=1$
$A=\int _{-2}^{\sqrt{6}}(2-(x^{2}-4)) d x$
$A=\int _{-2}^{\sqrt{6}}(6-x^{2}) d x=(6 x-\frac{x^{3}}{3}) _{-2}^{\sqrt{6}}$
$A=(6 \sqrt{6}-\frac{6 \sqrt{6}}{3})-(-12+\frac{8}{3})$
$A=\frac{12 \sqrt{6}}{3}+\frac{28}{3}$
$A=4 \sqrt{6}+\frac{28}{3}$
