If the area of the region bounded by 16x^2 - 9y^2 = 144 and 8x - 3y = 24 is A, then 3(A + …

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: 24

Accepted Answer

The hyperbola is $16x^2 - 9y^2 = 144$, and the line is $y = \dfrac{8x-24}{3}$.

Substituting into the hyperbola:
$$16x^2 - (8x-24)^2 = 144 \implies x^2 - 8x + 15 = 0 \implies x = 3, 5$$

For $3 \le x \le 5$, the upper hyperbola branch is $y = \dfrac{4}{3}\sqrt{x^2-9}$.

$$A = \int_3^5 \frac{4}{3}\sqrt{x^2-9}\,dx - \int_3^5 \frac{8x-24}{3}\,dx$$

Using $\displaystyle\int \sqrt{x^2-9}\,dx = \frac{x}{2}\sqrt{x^2-9} - \frac{9}{2}\log_e\left(x+\sqrt{x^2-9}\right)$:

$$\frac{4}{3}\left[\frac{x}{2}\sqrt{x^2-9} - \frac{9}{2}\log_e(x+\sqrt{x^2-9})\right]_3^5 = \frac{4}{3}\left(10 - \frac{9}{2}\log_e 9 + \frac{9}{2}\log_e 3\right) = \frac{40}{3} - 6\log_e 3$$

The line integral: $\displaystyle\int_3^5 \frac{8x-24}{3}\,dx = \frac{16}{3}$.

$$A = \frac{40}{3} - 6\log_e 3 - \frac{16}{3} = 8 - 6\log_e 3$$

So $A + 6\log_e 3 = 8$, and $3(A+6\log_e 3) = 24$.

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