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 \_\_\_\_\_\_\_\_.
Answer: 5
Solutions · 1
Official solution
The line meets the y-axis at (0,2). Solving the line and parabola simultaneously:
\[(y-2)^2+1=2y-4 \;\Rightarrow\; y^2-6y+9=0 \;\Rightarrow\; (y-3)^2=0 \;\Rightarrow\; y=3,\]
a repeated root --- the line is tangent to the parabola at (2,3). The region enclosed by the axes and the line up to this point of tangency:
\[\text{Area}=\int_0^2 \frac{x+4}{2}\,dx=\left[\frac{x^2}{4}+2x\right]_0^2=1+4=5.\]