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

The Correct Option is C

To solve for the area, we first correct the hyperbola equation and determine the bounds for integration. Transforming the hyperbola and solving for its axes involves completing the square and adjusting terms to standard form. Integration over the specified region, combined with constraints from the parabola, provides the desired area. The calculations involve determining intersections and integrating the correct function over the specified limits.

Given Equation:

\[ 16(x^2 + 4x) - (y^2 - 4y) + 44 = 0 \] 

Completing the Square:

\[ 16(x + 2)^2 - 64 - (y - 2)^2 + 4 + 44 = 0 \] \[ 16(x + 2)^2 - (y - 2)^2 = 16 \]

Transforming to Standard Form:

\[ \frac{(x + 2)^2}{1} - \frac{(y - 2)^2}{16} = 1 \]

Area Under the Curve:

\[ A = \int_{-2}^{\sqrt{6}} \left(6 - x^2\right) \, dx \] \[ A = \left[6x - \frac{x^3}{3}\right]_{-2}^{\sqrt{6}} \]

Calculations:

• At the upper limit (\(\sqrt{6}\)): \[ 6\sqrt{6} - \frac{(\sqrt{6})^3}{3} \]
• At the lower limit (\(-2\)): \[ 6(-2) - \frac{(-2)^3}{3} \]
• Subtracting the lower limit from the upper limit: \[ A = 12\sqrt{6} - \frac{6\sqrt{6}}{3} - \left(-12 + \frac{-8}{3}\right) \]

Simplifying:

\[ A = 4\sqrt{6} + \frac{28}{3} \]

Final Answer:

\[ A = 4\sqrt{6} + \frac{28}{3} \]

Answer 2

The correct answer is (C) : $4 \sqrt{6}+\frac{28}{3}$
$16\left(x^2+4x\right)-\left(y^2-4y\right)+44=0$
$16(x+2{)}^2-64-(y-2{)}^2+4+44=0$
$16(x+2{)}^2-(y-2{)}^2=16$
$\frac{(x+2{)}^2}{1}-\frac{(y-2{)}^2}{16}=1$

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