JEE Main 2025 — 3 April, Evening Shift — Question 41

Given $\frac{(x-6)^{2}}{a^{2}}-\frac{(y-2)^{2}}{b^{2}}=1$

$\Rightarrow b^{2} x^{2}-a^{2} y^{2}-12 x b^{2}+4 y a^{2}+36 b^{2}-4 a^{2}-a^{2} b^{2}=0$ Comparing $\frac{b^{2}}{a^{2}}=3 \Rightarrow e^{2}=1+\frac{b^{2}}{a^{2}}$

$$\begin{aligned} & \Rightarrow e^{2}=4 \\& \Rightarrow e=2 \end{aligned}$$

Similarly, $2 a e=4$

$\Rightarrow a=1 \Rightarrow b=\sqrt{3}$

$$\frac{(x-6)^{2}}{1}-\frac{(y-2)^{2}}{3}=1$$

$\Rightarrow 3 x^{2}-y^{2}-36 x+4 y+108-4-3=0$

$\Rightarrow 3 x^{2}-y^{2}-36 x+4 y+101=0$

$\Rightarrow \alpha=36, \beta=4, \gamma=101$

$\Rightarrow \alpha+\beta+\gamma=141$