JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
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 _____.

- A 140
- B 141
- C 142
- D 143
Answer & Solution
Correct Answer
(B) 141
Step-by-step Solution
Detailed explanation
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…
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