The point $P$ lies on the curve $y = \ln\left(\frac{x + 1}{x - 1}\right)$ for $x 1$.

Question

The point $P$ lies on the curve $y = \ln\left(\frac{x + 1}{x - 1}\right)$ for $x > 1$. The normal to the curve at $P$ is parallel to the line $2y = 3x + 2$. The equation of the tangent to the curve at $P$ cuts the $y$-axis at $Q$. Find the area of the triangle $POQ$, where $O$ is the origin.

Accepted Answer

Equation of tangent: $y = -\frac{2}{3}x + \ln{3} + \frac{4}{3}$ Area $=2.34$ units$^2$

2024 GESS PRELIMS P1 Q7

2024 GESS PRELIMS P1 Q7