JEE Main 2025 Apr 3 Shift 2, Mathematics Q25: Introduction to Hyperbola

JEE Main2025Apr 3, Shift 2Mathematics

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 ______.

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Solution

The centre of the hyperbola is the midpoint of the foci: $C = (6, 2)$ . The transverse axis is horizontal, half-distance between foci $c = 2$ .

The hyperbola has form $\frac{( x - 6 ) ^{2}}{a ^{2}} - \frac{( y - 2 ) ^{2}}{b ^{2}} = 1$ with $a^{2} + b^{2} = c^{2} = 4$ .

Clearing denominators: $b^{2} (x - 6)^{2} - a^{2} (y - 2)^{2} = a^{2} b^{2}$ .

Comparing the $x^{2}$ and $y^{2}$ coefficients with $3 x^{2} - y^{2}$ : the ratio of $x^{2}$ -coefficient to $y^{2}$ -coefficient should match. So $\frac{b ^{2}}{- a ^{2}} = \frac{3}{- 1} \Rightarrow b^{2} = 3 a^{2}$ .

Combined with $a^{2} + b^{2} = 4$ : $4 a^{2} = 4 \Rightarrow a^{2} = 1, b^{2} = 3$ .

Now expand $3 (x - 6)^{2} - (y - 2)^{2} = 3$ :

$3 x^{2} - 36 x + 108 - y^{2} + 4 y - 4 = 3$

$3 x^{2} - y^{2} - 36 x + 4 y + 101 = 0$ .

Identify: $α = 36, β = 4, γ = 101$ .

$α + β + γ = 36 + 4 + 101 = 141$ .

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JEE Main 2025 Apr 3 Shift 2, Mathematics Q25: Introduction to Hyperbola

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