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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 _____.

  1. A 140
  2. B 141
  3. C 142
  4. D 143
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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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