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JEE Main 20253 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
3 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Hyperbola

JEE Main 2025 · 3 April · Evening Shift · Mathematics · Hyperbola

If the equation of the hyperbola with foci (4,2) and (8,2) is 3...

If the equation of the hyperbola with foci (4,2)(4,2) and (8,2)(8,2) is 3x2−y2−αx+βy+γ=03 x^2-y^2-\alpha x+\beta y+\gamma=0, then α+β+γ\alpha+\beta+\gamma is equal to _____.
If the equation of the hyperbola with foci (4,2) and (8,2) is 3 x^2-y^2-α x+β y+γ=0 , then α+β+γ is equal to _____.

Correct answer

0

Step-by-step solution

Equation of hyperbola is
(x−6)2a2−(y−2)24−a2=1\frac{(x-6)^2}{a^2}-\frac{(y-2)^2}{4-a^2}=1
⇒(4−a2)(x−6)2−a2(y−2)2=a2(4−a2)\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 3x2−y2−αx+βy+γ=03 x^2-y^2-\alpha x+\beta y+\gamma=0, we get a2=1\mathrm{a}^2=1 and α=36,β=4\alpha=36, \beta=4 and γ=101\gamma=101
∴α+β+γ=141\therefore \alpha+\beta+\gamma=141
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