JEE MAIN - Mathematics (2025 - 3rd April Evening Shift - No. 22)
If the equation of the hyperbola with foci $(4,2)$ and $(8,2)$ is $3 x^2-y^2-\alpha x+\beta y+\gamma=0$, then $\alpha+\beta+\gamma$ is equal to__________.
Answer
141
Explanation
$$\begin{aligned} & \text { Given } \frac{(x-6)^2}{a^2}-\frac{(y-2)^2}{b^2}=1 \\ & \begin{aligned} & \Rightarrow b^2 x^2-a^2 y^2-12 x b^2+4 y a^2+36 b^2-4 a^2-a^2 b^2=0 \\ & \text { Comparing } \frac{b^2}{a^2}=3 \Rightarrow e^2=1+\frac{b^2}{a^2} \\ & \Rightarrow e^2=4 \\ & \Rightarrow e=2 \end{aligned} \end{aligned}$$
$$\begin{aligned} &\text { Similarly, } 2 a e=4\\ &\begin{aligned} \Rightarrow & a=1 \Rightarrow b=\sqrt{3} \\ & \frac{(x-6)^2}{1}-\frac{(y-2)^2}{3}=1 \\ \Rightarrow & 3 x^2-y^2-36 x+4 y+108-4-3=0 \\ \Rightarrow & 3 x^2-y^2-36 x+4 y+101=0 \\ \Rightarrow & \alpha=36, \beta=4, \gamma=101 \\ \Rightarrow & \alpha+\beta+\gamma=141 \end{aligned} \end{aligned}$$
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