Area Under the Curve
Area Between Hyperbola and Line
nta_pyq_2024_jan
Grade 12
Question:
The area enclosed by the curves $xy+4y=16$ and $x+y=6$ is equal to:
$28-30\log_e 2$
$30-28\log_e 2$
$30-32\log_e 2$
$32-30\log_e 2$
Step-by-Step Solution
Key Concept: Rewrite hyperbola as $y=\frac{16}{x+4}$. Find intersections with $y=6-x$: solve $\frac{16}{x+4}=6-x$ giving $x=-2$ and $x=4$. Integrate $(6-x)-\frac{16}{x+4}$ from $-2$ to $4$.
Intersections at $x=-2$ and $x=4$. Area $=\int_{-2}^{4}\left[(6-x)-\frac{16}{x+4}\right]dx=\left[6x-\frac{x^2}{2}-16\ln|x+4|\right]_{-2}^{4}=30-32\ln 2$.
Correct Answer: 3