If the equation of the hyperbola with foci (4, 2) and (8, 2) is , then is equal to __________. [2025]

Question

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 __________. [2025]

Smart Achievers · Accepted Answer

(141)

Equation of hyperbola is $\frac{{(x - 6)}^{2}}{a^{2}} - \frac{{(y - 2)}^{2}}{4 - a^{2}} = 1$

$\Rightarrow (4 - a^{2}) {(x - 6)}^{2} - a^{2} {(y - 2)}^{2} = a^{2} (4 - a^{2})$

$\Rightarrow (4 - a^{2}) x^{2} - a^{2} y^{2} + (12 a^{2} - 48) x + 4 a^{2} y + 144 - 44 a^{2} + a^{4} = 0$

On comparing with $3 x^{2} - y^{2} - α x + β y + γ = 0$ , we get $a^{2} = 1$ and $α = 36$ , $β = 4$ and $γ = 101$

$∴ α + β + γ = 141$

Solution

Q.
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 __________. [2025]

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Solution

Q. If the equation of the hyperbola with foci (4, 2) and (8, 2) is 3x2–y2–αx+βy+γ=0, then α+β+γ is equal to __________. [2025]

Ans. (141) Equation of hyperbola is (x–6)2a2–(y–2)24–a2=1 ⇒ (4–a2)(x–6)2–a2(y–2)2=a2(4–a2) ⇒ (4–a2)x2–a2y2+(12a2–48)x+4a2y+144–44a2+a4=0 On comparing with 3x2–y2–αx+βy+γ=0, we get a2=1 and α=36, β=4 and γ=101 ∴ α+β+γ=141

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Q.
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 __________. [2025]