JEE Main 2024 (Online) 30th January Evening ShiftQuestion 1
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 $\_\_\_\_\_\_$ .
Application of Integrals
Answer$5$
Solution
Parabola $x=1+(y-2)^2$ and line $x=2y-4$ meet where $(y-3)^2=0$, i.e. tangent at $(2,3)$. Parabola meets x-axis at $(5,0)$. Region: $0\le y\le3$, $x$ from $\max(0,2y-4)$ to $1+(y-2)^2$. Area $=\int_0^3(1+(y-2)^2)dy-\int_2^3(2y-4)dy=(3+3)-1=5$.