If the equation of the hyperbola with foci (4 | vedclass

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(A) The center of the hyperbola is the midpoint of the foci $(4,2)$ and $(8,2)$,which is $C = (\frac{4+8}{2}, \frac{2+2}{2}) = (6,2)$.The distance bet

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$141$

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(A) The center of the hyperbola is the midpoint of the foci $(4,2)$ and $(8,2)$,which is $C = (\frac{4+8}{2}, \frac{2+2}{2}) = (6,2)$.The distance between the foci is $2ae = 8-4 = 4$,so $ae = 2$.The equation of the hyperbola is $\frac{(x-6)^2}{a^2} - \frac{(y-2)^2}{b^2} = 1$,where $b^2 = a^2(e^2-1) = a^2e^2 - a^2 = 4 - a^2$.Given the equation $3x^2 - y^2 - \alpha x + \beta y + \gamma = 0$,the coefficients of $x^2$ and $y^2$ imply $\frac{1}{a^2} = 3$ and $\frac{1}{b^2} = 1$,so $a^2 = \frac{1}{3}$ and $b^2 = 1$.However,$b^2 = 4 - a^2$ gives $1 = 4 - a^2$,so $a^2 = 3$ and $b^2 = 1$.Substituting $a^2=3$ and $b^2=1$ into the standard form: $\frac{(x-6)^2}{3} - \frac{(y-2)^2}{1} = 1$.$(x-6)^2 - 3(y-2)^2 = 3 \Rightarrow (x^2 - 12x + 36) - 3(y^2 - 4y + 4) = 3$.$x^2 - 12x + 36 - 3y^2 + 12y - 12 = 3 \Rightarrow x^2 - 3y^2 - 12x + 12y + 21 = 0$.Comparing this with $3x^2 - y^2 - \alpha x + \beta y + \gamma = 0$ is not directly possible as the coefficients of $x^2$ and $y^2$ are different. Re-evaluating the given equation $3x^2 - y^2 - \alpha x + \beta y + \gamma = 0$,we have $3(x^2 - \frac{\alpha}{3}x) - (y^2 - \beta y) + \gamma = 0$.$3(x - \frac{\alpha}{6})^2 - (y - \frac{\beta}{2})^2 = -\gamma + \frac{\alpha^2}{12} - \frac{\beta^2}{4}$.Center is $(\frac{\alpha}{6}, \frac{\beta}{2}) = (6,2) \Rightarrow \alpha = 36, \beta = 4$.Equation: $3(x-6)^2 - (y-2)^2 = -\gamma + \frac{36^2}{12} - \frac{4^2}{4} = -\gamma + 108 - 4 = 104 - \gamma$.For this to be a hyperbola with $a^2=1, b^2=3$,the $RHS$ must be $3$. So $104 - \gamma = 3 \Rightarrow \gamma = 101$.Thus,$\alpha + \beta + \gamma = 36 + 4 + 101 = 141$.

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

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