JEE Main 2025 — Mathematics Coordinate Geometry

$r=\left|\frac{a+1}{\sqrt{2}}\right| \Rightarrow(a+1)^2=2 r^2$r=​2​a+1​​⇒(a+1)2=2r2 Also $\left(\frac{3 a-1}{\sqrt{13}}\right)^2+\left(\frac{2}{\sqrt{13}}\right)^2=r^2$(13​3a−1​)2+(13​2​)2=r2 $\begin{gathered}
\Rightarrow\left(\frac{3 a-1}{\sqrt{13}}\right)^2+\frac{4}{13}=\frac{(a+1)^2}{2} \
5 a^2-14 a-3=0
\end{gathered}$$\therefore \quad a=-\frac{1}{5}, 3$$\begin{aligned}
& \because \quad a \neq-\frac{1}{5} \Rightarrow \
& \Rightarrow \quad r=2 \sqrt{2}
\end{aligned}$$\because$One focus of$Onefocusof\frac{x^2}{\alpha^2}-\frac{y^2}{\beta^2}=1$is$is(3,0)$$\begin{aligned} & \Rightarrow \alpha e=3 \text { and } 2 \alpha=4 \sqrt{2} \ & \Rightarrow \alpha=2 \sqrt{2} \Rightarrow \alpha^2=8 \ & \alpha^2\left[1+\frac{\beta^2}{\alpha^2}\right]=9 \ & \alpha^2+\beta^2=9 \ & \Rightarrow \beta^2=1 \ & \therefore 2 \alpha^2+3 \beta^2=19\end{aligned}$