Area Under the Curve
Area Enclosed by Parabola, Line, and Axes
nta_pyq_2024_jan
Grade 12
Question:
The area of the region enclosed by the parabola $(y-2)^2=x-1$, the line $x-2y+4=0$ and the positive coordinate axes is
Step-by-Step Solution
Key Concept: The line is $x=2y-4$. The parabola vertex is at $(1,2)$. Find intersections: $(y-2)^2=2y-4-1=2y-5$; let $u=y-2$: $u^2=2u-1\Rightarrow(u-1)^2=0\Rightarrow y=3$, $x=2$. They are tangent at $(2,3)$. Compute area enclosed with coordinate axes.
Area $=\int_0^3[(y-2)^2+1]dy-\frac{1}{2}\times1\times2=\left[\frac{(y-2)^3}{3}+y\right]_0^3-1=(\frac{1}{3}+3)-(\frac{-8}{3}+0)-1=9-18+15-1=5$.
Correct Answer: 5