JEE Main 2024 — Mathematics Calculus

Given: ${x}^{2}+{y}^{2}=169,5x-y=13$x2+y2=169,5x−y=13

$$\Rightarrow {x}^{2}+{(5x-13)}^{2}=169$$⇒x2+(5x−13)2=169

$$\Rightarrow {x}^{2}+25{x}^{2}+169-130x-169=0$$⇒x2+25x2+169−130x−169=0

$$\Rightarrow 26{x}^{2}-130x=0$$⇒26x2−130x=0

$$\Rightarrow x(x-5)=0$$⇒x(x−5)=0

$$\Rightarrow x=0,5$$⇒x=0,5

$$\Rightarrow y=-13,12$$⇒y=−13,12

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

So, the required area is given by,

$$A={\int }_{-13}^{12}\sqrt{169-{y}^{2}}\mathrm{dy}-\frac{1}{2}\times 25\times 5$$A=∫−1312​169−y2​dy−21​×25×5

$$\Rightarrow A={[\frac{x}{2}\sqrt{169-{x}^{2}}+\frac{169}{2}{\mathrm{sin}}^{-1}\frac{x}{13}]}_{-13}^{12}-\frac{1}{2}\times 25\times 5$$⇒A=[2x​169−x2​+2169​sin−113x​]−1312​−21​×25×5

$$\Rightarrow A=\frac{\pi }{2}\times \frac{169}{2}-\frac{65}{2}+\frac{169}{2}{\mathrm{sin}}^{-1}\frac{12}{13}$$⇒A=2π​×2169​−265​+2169​sin−11312​

Hence, on comparing with given value we get,

$$\Rightarrow \alpha =169,\beta =2$$⇒α=169,β=2

$$\Rightarrow \alpha +\beta =171$$⇒α+β=171