Question

If the area bounded by two curves \( \dfrac{x^2}{9} - \dfrac{y^2}{16} = 1 \) and \( 8x - 3y = 24 \) is \( A - 6\log_e 3 \), then \( A \) is equal to

• A. 5
• B. 6
• C. 7
• D. 8

Hint:

When area is bounded by a curve and a line, first find the intersection points, then identify which graph lies above the other in the required interval before setting up the integral.

Answer 1

The Correct Option is D

Step 1: Find the points of intersection of the line and the hyperbola.
Given line is
\[ 8x - 3y = 24 \] So,
\[ y = \frac{8x-24}{3}. \] Now substitute this in the hyperbola equation
\[ \frac{x^2}{9} - \frac{y^2}{16} = 1. \] We get
\[ \frac{x^2}{9} - \frac{1}{16}\left(\frac{8x-24}{3}\right)^2 = 1. \] Multiplying by \( 9 \),
\[ x^2 - \frac{(8x-24)^2}{16} = 9. \] Now,
\[ \frac{(8x-24)^2}{16} = 4(x-3)^2. \] Hence,
\[ x^2 - 4(x-3)^2 = 9. \] Expanding,
\[ x^2 - 4(x^2-6x+9) = 9 \] \[ x^2 - 4x^2 + 24x - 36 = 9 \] \[ -3x^2 + 24x - 45 = 0 \] \[ 3x^2 - 24x + 45 = 0 \] \[ x^2 - 8x + 15 = 0 \] \[ (x-3)(x-5)=0. \] So,
\[ x=3 \quad \text{or} \quad x=5. \] Now corresponding \( y \)-values are
for \( x=3 \),
\[ y=\frac{24-24}{3}=0, \] and for \( x=5 \),
\[ y=\frac{40-24}{3}=\frac{16}{3}. \] Thus, the curves intersect at
\[ (3,0) \quad \text{and} \quad \left(5,\frac{16}{3}\right). \]
Step 2: Write the equations of the upper branch of the hyperbola and the line.
From the hyperbola,
\[ \frac{x^2}{9} - \frac{y^2}{16} = 1 \] we get
\[ \frac{y^2}{16} = \frac{x^2}{9} - 1 \] \[ y = \frac{4}{3}\sqrt{x^2-9}. \] The line is
\[ y=\frac{8x-24}{3}. \] Between \( x=3 \) and \( x=5 \), the upper branch of the hyperbola lies above the line.
Hence required area is
\[ \int_{3}^{5} \left( \frac{4}{3}\sqrt{x^2-9} - \frac{8x-24}{3} \right) dx. \]
Step 3: Evaluate the integral.
So,
\[ \text{Area}=\frac{4}{3}\int_{3}^{5}\sqrt{x^2-9}\,dx-\frac{1}{3}\int_{3}^{5}(8x-24)\,dx. \] Using the standard formula
\[ \int \sqrt{x^2-a^2}\,dx=\frac{1}{2}\left[x\sqrt{x^2-a^2}-a^2\ln\left(x+\sqrt{x^2-a^2}\right)\right], \] with \( a=3 \), we get
\[ \int \sqrt{x^2-9}\,dx=\frac{1}{2}\left[x\sqrt{x^2-9}-9\ln\left(x+\sqrt{x^2-9}\right)\right]. \] Therefore,
\[ \frac{4}{3}\int_{3}^{5}\sqrt{x^2-9}\,dx = \frac{2}{3}\left[x\sqrt{x^2-9}-9\ln\left(x+\sqrt{x^2-9}\right)\right]_{3}^{5}. \] At \( x=5 \),
\[ \sqrt{25-9}=4, \] so the value becomes
\[ 5\cdot 4 - 9\ln(5+4)=20-9\ln 9. \] At \( x=3 \),
\[ \sqrt{9-9}=0, \] so the value becomes
\[ 0-9\ln 3. \] Hence,
\[ \frac{2}{3}\left[(20-9\ln 9)-(-9\ln 3)\right] = \frac{2}{3}\left[20-9\ln 9+9\ln 3\right]. \] Since \( \ln 9 = 2\ln 3 \),
\[ 20-9\ln 9+9\ln 3 = 20-18\ln 3+9\ln 3 = 20-9\ln 3. \] Thus,
\[ \frac{4}{3}\int_{3}^{5}\sqrt{x^2-9}\,dx = \frac{2}{3}(20-9\ln 3) = \frac{40}{3}-6\ln 3. \] Now,
\[ \frac{1}{3}\int_{3}^{5}(8x-24)\,dx = \frac{1}{3}\left[4x^2-24x\right]_{3}^{5}. \] So,
\[ \frac{1}{3}\left[(100-120)-(36-72)\right] = \frac{1}{3}(-20+36) = \frac{16}{3}. \] Therefore, total area is
\[ \left(\frac{40}{3}-6\ln 3\right)-\frac{16}{3} = \frac{24}{3}-6\ln 3 = 8-6\ln 3. \]
Step 4: Compare with the given form.
Given area is
\[ A-6\log_e 3. \] Comparing,
\[ A-6\log_e 3 = 8-6\log_e 3. \] Hence,
\[ A=8. \]
Final Answer: \( 8 \)