JEE Main 2024 — Area Under Curves Question with Solution

Given: $x^2+y^2=169, 5x-y=13$

$\Rightarrow x^2+{\left(5x-13\right)}^2=169$

$\Rightarrow x^2+25x^2+169-130x-169=0$

$\Rightarrow 26x^2-130x=0$

$\Rightarrow x\left(x-5\right)=0$

$\Rightarrow x=0,5$

$\Rightarrow y=-13, 12$

So, the points of intersection are $\left(0,-13\right) \text{and} \left(5,12\right)$.

So, the required area is given by,

$A={\int }_{-13}^{12}\sqrt{169-y^2}\mathrm{dy}-\frac{1}{2}\times 25\times 5$

$\Rightarrow A={\left[\frac{x}{2}\sqrt{169-x^2}+\frac{169}{2}{\sin }^{-1}\frac{x}{13}\right]}_{-13}^{12}-\frac{1}{2}\times 25\times 5$

$\Rightarrow A=\frac{\pi }{2}\times \frac{169}{2}-\frac{65}{2}+\frac{169}{2}{\sin }^{-1}\frac{12}{13}$

Hence, on comparing with given value we get,

$\Rightarrow \alpha =169, \beta =2$

$\Rightarrow \alpha +\beta =171$