Area Bounded by Hyperbola and Straight Line

To find the area $A$ of the region bounded by the hyperbola $16x^2 - 9y^2 = 144$ and the straight line $8x - 3y = 24$, we first determine their points of intersection.

The equation of the hyperbola can be rewritten as: $\frac{x^2}{9} - \frac{y^2}{16} = 1$

From the equation of the line, we express $3y$ in terms of $x$: $3y = 8x - 24 = 8(x - 3)$

Substitute $3y = 8x - 24$ into the hyperbola equation $16x^2 - (3y)^2 = 144$: $16x^2 - (8x - 24)^2 = 144$ $16x^2 - (64x^2 - 384x + 576) = 144$ $-48x^2 + 384x - 720 = 0$

Dividing the entire equation by $-48$: $x^2 - 8x + 15 = 0$ $(x - 3)(x - 5) = 0$

Thus, the $x$-coordinates of the intersection points are $x = 3$ and $x = 5$.

For $x \in [3, 5]$ in the first quadrant, the upper boundary curve is the hyperbola: $y_1 = \frac{4}{3}\sqrt{x^2 - 9}$

and the lower boundary curve is the straight line: $y_2 = \frac{8}{3}(x - 3)$

The area $A$ of the bounded region is given by the integral: $A = \int_{3}^{5} (y_1 - y_2) \, dx = \int_{3}^{5} \left( \frac{4}{3}\sqrt{x^2 - 9} - \frac{8}{3}(x - 3) \right) dx$

We split this integral into two parts, $I_1$ and $I_2$:

1. Calculating $I_1 = \frac{4}{3} \int_{3}^{5} \sqrt{x^2 - 9} \, dx$: Using the standard integration formula $\int \sqrt{x^2 - a^2} \, dx = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\log_e\left|x + \sqrt{x^2 - a^2}\right|$:

$\int_{3}^{5} \sqrt{x^2 - 9} \, dx = \left[ \frac{x}{2}\sqrt{x^2 - 9} - \frac{9}{2}\log_e\left(x + \sqrt{x^2 - 9}\right) \right]_{3}^{5}$ $= \left( \frac{5}{2}\sqrt{25 - 9} - \frac{9}{2}\log_e(5 + 4) \right) - \left( 0 - \frac{9}{2}\log_e(3) \right)$ $= \left( 10 - \frac{9}{2}\log_e(9) \right) + \frac{9}{2}\log_e(3)$ $= 10 - 9\log_e(3) + \frac{9}{2}\log_e(3) = 10 - \frac{9}{2}\log_e(3)$

Thus: $I_1 = \frac{4}{3} \left( 10 - \frac{9}{2}\log_e(3) \right) = \frac{40}{3} - 6\log_e(3)$

1. Calculating $I_2 = \int_{3}^{5} \frac{8}{3}(x - 3) \, dx$: $I_2 = \frac{8}{3} \left[ \frac{(x - 3)^2}{2} \right]_{3}^{5} = \frac{8}{3} \cdot \frac{(5 - 3)^2}{2} = \frac{8}{3} \cdot 2 = \frac{16}{3}$

Now, combining the results for area $A$: $A = I_1 - I_2 = \left( \frac{40}{3} - 6\log_e(3) \right) - \frac{16}{3}$ $A = \frac{24}{3} - 6\log_e(3) = 8 - 6\log_e(3)$

Rearranging terms: $A + 6\log_e(3) = 8$

Multiplying by $3$: $3\left(A + 6\log_e(3)\right) = 3 \times 8 = 24$