Question

If the area of the region bounded by \(16x^2 - 9y^2 = 144\) and \(8x - 3y = 24\) is \(A\), then \(3(A + 6 \log_e(3))\) is equal to _______.

Answer 1

Correct Answer: 37

Step 1: Understanding the Concept:
We are dealing with the area between a hyperbola and a straight line.
First, we find the points of intersection by solving the two equations simultaneously.
Then, we set up a definite integral of the upper curve minus the lower curve between the intersection points.
Step 2: Key Formula or Approach:
The hyperbola is \(\frac{x^2}{9} - \frac{y^2}{16} = 1\).
The line is \(8x - 3y = 24 \implies y = \frac{8}{3}x - 8\).
Substitute \(y\) from the line into the hyperbola equation to find intersection limits \([x_1, x_2]\).
Area \(A = \int_{x_1}^{x_2} (y_{\text{upper}} - y_{\text{lower}}) dx\).
Standard integral: \(\int \sqrt{x^2 - a^2} dx = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\ln|x + \sqrt{x^2 - a^2}|\).
Step 3: Detailed Explanation:
Find the intersection points:
\[ 16x^2 - 9\left(\frac{8}{3}x - 8\right)^2 = 144 \] \[ 16x^2 - 9\left(\frac{64}{9}x^2 - \frac{128}{3}x + 64\right) = 144 \] \[ 16x^2 - 64x^2 + 384x - 576 = 144 \] \[ -48x^2 + 384x - 720 = 0 \] Divide by \(-48\):
\[ x^2 - 8x + 15 = 0 \implies (x-3)(x-5) = 0 \] The intersection points are at \(x = 3\) and \(x = 5\).
In the interval \([3, 5]\), let's check which curve is higher. At \(x=4\):
Line: \(y = \frac{8}{3}(4) - 8 = \frac{32}{3} - \frac{24}{3} = \frac{8}{3} \approx 2.67\)
Hyperbola: \(y = 4\sqrt{\frac{4^2}{9} - 1} = \frac{4\sqrt{7}}{3} \approx \frac{4(2.64)}{3} \approx 3.52\)
Thus, the hyperbola is above the line.
\[ y_{\text{hyp}} = 4\sqrt{\frac{x^2}{9} - 1} = \frac{4}{3}\sqrt{x^2 - 9} \] \[ A = \int_{3}^{5} \left( \frac{4}{3}\sqrt{x^2 - 9} - \left(\frac{8}{3}x - 8\right) \right) dx \] Integrate the linear part first:
\[ \int_{3}^{5} \left(\frac{8}{3}x - 8\right) dx = \left[ \frac{4}{3}x^2 - 8x \right]_{3}^{5} \] \[ = \left( \frac{100}{3} - 40 \right) - \left( \frac{36}{3} - 24 \right) = \left( -\frac{20}{3} \right) - (-12) = 12 - \frac{20}{3} = \frac{16}{3} \] Now integrate the hyperbola part:
\[ \int_{3}^{5} \frac{4}{3}\sqrt{x^2 - 9} dx = \frac{4}{3} \left[ \frac{x}{2}\sqrt{x^2 - 9} - \frac{9}{2}\ln|x + \sqrt{x^2 - 9}| \right]_{3}^{5} \] \[ = \frac{4}{3} \left[ \left( \frac{5}{2}(4) - \frac{9}{2}\ln(5+4) \right) - \left( 0 - \frac{9}{2}\ln(3) \right) \right] \] \[ = \frac{4}{3} \left[ 10 - \frac{9}{2}\ln 9 + \frac{9}{2}\ln 3 \right] = \frac{4}{3} \left[ 10 - 9\ln 3 + \frac{9}{2}\ln 3 \right] = \frac{4}{3} \left[ 10 - \frac{9}{2}\ln 3 \right] = \frac{40}{3} - 6\ln 3 \] Subtract the linear part from the hyperbola part to get the area:
\[ A = \left( \frac{40}{3} - 6\ln 3 \right) - \frac{16}{3} = \frac{24}{3} - 6\ln 3 = 8 - 6\ln 3 \] We need to evaluate \(3(A + 6\ln 3)\).
\[ 3(8 - 6\ln 3 + 6\ln 3) = 3(8) = 24 \] Step 4: Final Answer:
The value is \(24\).