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 _____.

Question

Aicharya · Accepted Answer

Equation of hyperbola is
$\frac{(x-6)^2}{a^2}-\frac{(y-2)^2}{4-a^2}=1$
$\Rightarrow\left(4-\mathrm{a}^2\right)(\mathrm{x}-6)^2-\mathrm{a}^2(\mathrm{y}-2)^2=\mathrm{a}^2\left(4-\mathrm{a}^2\right)$
comparing with $3 x^2-y^2-\alpha x+\beta y+\gamma=0$, we get $\mathrm{a}^2=1$ and $\alpha=36, \beta=4$ and $\gamma=101$
$\therefore \alpha+\beta+\gamma=141$

If the equation of the hyperbola with foci $(4,2)$ and $(8,2)$ is $3 x^2-y^2-\alpha x+\beta y+\gamma=0$,…

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