Q118 / 160 · TS EAMCET 2024 (10 May Shift 1) · Mathematics
The area of the region enclosed by the curves $y^2=4(x+1)$ and $y^2=5(x-4)$ is
- A280/3
- B150
- C140
- Correct answerD200/3
Solution
$y^2=4(x+1)$ and $y^2=5(x-4)$
Point of intersections are $(24,10)$ and $(24,-10)$
$\begin{aligned} & \text { Required area }=\int_{-10}^{10}(\text { Upper }- \text { Lower) Curve } \\ & =\int_{-10}^{10}\left(\frac{y^2+20}{5}-\frac{y^2-4}{5}\right) d y=\int_{-10}^{10} \frac{-y^2+100}{20} d y \\ & =\frac{1}{10}\left[\frac{-y^3}{30}+100 y\right]_0^{10}=\frac{200}{3}\end{aligned}$
$\begin{aligned} & \text { Required area }=\int_{-10}^{10}(\text { Upper }- \text { Lower) Curve } \\ & =\int_{-10}^{10}\left(\frac{y^2+20}{5}-\frac{y^2-4}{5}\right) d y=\int_{-10}^{10} \frac{-y^2+100}{20} d y \\ & =\frac{1}{10}\left[\frac{-y^3}{30}+100 y\right]_0^{10}=\frac{200}{3}\end{aligned}$
Hence, the correct answer is option: 'D'.