Hyperbola and Line Equations

The given equations are:

• Hyperbola: $16x^2 - 9y^2 = 144$. Dividing by 144 gives the standard form: $\frac{x^2}{9} - \frac{y^2}{16} = 1$.
• Line: $8x - 3y = 24$. Solving for $y$, we get $3y = 8x - 24$, which simplifies to $y = \frac{8}{3}x - 8$.

Intersection Points

To find the points where the line intersects the hyperbola, substitute the expression for $y$ from the line equation into the hyperbola equation:

$16x^2 - 9\left(\frac{8}{3}x - 8\right)^2 = 144$

$16x^2 - 9\left(\frac{64}{9}x^2 - \frac{128}{3}x + 64\right) = 144$

$16x^2 - 64x^2 + 384x - 576 = 144$

$-48x^2 + 384x - 720 = 0$

Divide by -48: $x^2 - 8x + 15 = 0$

Factor the quadratic: $(x-3)(x-5) = 0$. The solutions are $x=3$ and $x=5$.

Find the corresponding $y$ values using $y = \frac{8}{3}x - 8$:

• For $x=3$, $y = \frac{8}{3}(3) - 8 = 8 - 8 = 0$. Point: $(3, 0)$.
• For $x=5$, $y = \frac{8}{3}(5) - 8 = \frac{40}{3} - \frac{24}{3} = \frac{16}{3}$. Point: $(5, \frac{16}{3})$.

Area Calculation

The area $A$ is the region bounded by the hyperbola and the line segment between the intersection points. The upper branch of the hyperbola is given by $y = \frac{4}{3}\sqrt{x^2-9}$ (since $y \geq 0$ between the intersection points). We need to determine if the hyperbola or the line is above the other in the interval $[3, 5]$.

Test $x=4$: Line $y = \frac{8}{3}(4) - 8 = \frac{8}{3}$. Hyperbola $y = \frac{4}{3}\sqrt{4^2-9} = \frac{4}{3}\sqrt{7}$. Since $\sqrt{7} > 2$, the hyperbola is above the line.

The area $A$ is calculated by the integral:

$A = \int_{3}^{5} \left( \frac{4}{3}\sqrt{x^2-9} - \left(\frac{8}{3}x - 8\right) \right) dx$

$A = \int_{3}^{5} \frac{4}{3}\sqrt{x^2-9} dx - \int_{3}^{5} \left(\frac{8}{3}x - 8\right) dx$

Evaluate the first integral using the formula $\int \sqrt{x^2 - a^2} dx = \frac{x}{2}\sqrt{x^2-a^2} - \frac{a^2}{2}\log_e|x + \sqrt{x^2-a^2}|$, with $a=3$:

$\int_{3}^{5} \frac{4}{3}\sqrt{x^2-9} dx = \frac{4}{3} \left[ \frac{x}{2}\sqrt{x^2-9} - \frac{9}{2}\log_e|x + \sqrt{x^2-9}| \right]_{3}^{5}$

= $\frac{4}{3} \left[ \left( \frac{5}{2}(4) - \frac{9}{2}\log_e(9) \right) - \left( 0 - \frac{9}{2}\log_e(3) \right) \right]$

= $\frac{4}{3} \left[ 10 - \frac{9}{2}(2\log_e(3)) + \frac{9}{2}\log_e(3) \right]$

= $\frac{4}{3} \left[ 10 - 9\log_e(3) + \frac{9}{2}\log_e(3) \right]$

= $\frac{4}{3} \left[ 10 - \frac{9}{2}\log_e(3) \right] = \frac{40}{3} - 6\log_e(3)$

Evaluate the second integral:

$\int_{3}^{5} \left(\frac{8}{3}x - 8\right) dx = \left[ \frac{4}{3}x^2 - 8x \right]_{3}^{5}$

= $\left( \frac{4}{3}(25) - 40 \right) - \left( \frac{4}{3}(9) - 24 \right)$

= $\left( \frac{100}{3} - \frac{120}{3} \right) - (12 - 24) = -\frac{20}{3} - (-12) = -\frac{20}{3} + \frac{36}{3} = \frac{16}{3}$

Combine the results to find $A$:

$A = \left(\frac{40}{3} - 6\log_e(3)\right) - \frac{16}{3} = \frac{24}{3} - 6\log_e(3) = 8 - 6\log_e(3)$

Final Expression Value

We need to calculate $3(A + 6\log_e(3))$:

$3(A + 6\log_e(3)) = 3((8 - 6\log_e(3)) + 6\log_e(3))$

$= 3(8)$

= $24$