JEE Mathematics 2025 Question with Solution

Given: The foci are $(4, 2)$ and $(8, 2)$, and the hyperbola is $3x^2 - y^2 - \alpha x + \beta y + \gamma = 0$.

Find: The value of $\alpha + \beta + \gamma$.

The center is the midpoint of the foci:

$$h = \frac{4+8}{2} = 6, \qquad k = \frac{2+2}{2} = 2$$

So the center is $(6,2)$.

The distance between the foci is $2c = 4$, hence

$$c = 2 \implies c^2 = 4$$

Since the y-coordinates of the foci are the same, the transverse axis is horizontal.

Now complete the square in the given equation:

$$(3x^2 - \alpha x) - (y^2 - \beta y) + \gamma = 0$$ $$3\left(x^2 - \frac{\alpha}{3}x\right) - \left(y^2 - \beta y\right) + \gamma = 0$$ $$3\left(x^2 - \frac{\alpha}{3}x + \left(\frac{\alpha}{6}\right)^2\right) - 3\left(\frac{\alpha}{6}\right)^2 - \left(y^2 - \beta y + \left(\frac{\beta}{2}\right)^2\right) + \left(\frac{\beta}{2}\right)^2 + \gamma = 0$$ $$3\left(x - \frac{\alpha}{6}\right)^2 - \left(y - \frac{\beta}{2}\right)^2 = \frac{\alpha^2}{12} - \frac{\beta^2}{4} - \gamma$$

Let

$$K = \frac{\alpha^2}{12} - \frac{\beta^2}{4} - \gamma$$

Then the equation becomes

$$\frac{\left(x - \alpha/6\right)^2}{K/3} - \frac{\left(y - \beta/2\right)^2}{K} = 1$$

Comparing the center with $(6,2)$,

$$\frac{\alpha}{6} = 6 \implies \alpha = 36$$ $$\frac{\beta}{2} = 2 \implies \beta = 4$$

Also,

$$a^2 = \frac{K}{3}, \qquad b^2 = K$$

Using the hyperbola relation

$$c^2 = a^2 + b^2$$

we get

$$4 = \frac{K}{3} + K = \frac{4K}{3}$$ $$K = 3$$

Now substitute into

$$K = \frac{\alpha^2}{12} - \frac{\beta^2}{4} - \gamma$$ $$3 = \frac{36^2}{12} - \frac{4^2}{4} - \gamma$$ $$3 = \frac{1296}{12} - \frac{16}{4} - \gamma = 108 - 4 - \gamma$$ $$3 = 104 - \gamma \implies \gamma = 101$$

Therefore,

$$\alpha + \beta + \gamma = 36 + 4 + 101 = 141$$

So, the value of $\alpha + \beta + \gamma$ is $141$.

Given: The foci are $(4,2)$ and $(8,2)$.

Find: The value of $\alpha + \beta + \gamma$.

The center is the midpoint:

$$\left(\frac{4+8}{2}, \frac{2+2}{2}\right) = (6,2)$$

Also,

$$2c = 4 \implies c = 2$$

Since the foci have the same y-coordinate, the hyperbola is

$$\frac{(x-6)^2}{a^2} - \frac{(y-2)^2}{b^2} = 1$$

with

$$c^2 = a^2 + b^2 = 4$$

Now expand the standard equation:

$$b^2(x-6)^2 - a^2(y-2)^2 = a^2b^2$$ $$b^2(x^2-12x+36) - a^2(y^2-4y+4) = a^2b^2$$ $$b^2x^2 - a^2y^2 - 12b^2x + 4a^2y + 36b^2 - 4a^2 - a^2b^2 = 0$$

Compare this with

$$3x^2 - y^2 - \alpha x + \beta y + \gamma = 0$$

This gives a scaling comparison:

$$b^2 : a^2 = 3 : 1$$

So,

$$b^2 = 3a^2$$

Using

$$a^2 + b^2 = 4$$

we get

$$a^2 + 3a^2 = 4$$ $$4a^2 = 4 \implies a^2 = 1, \qquad b^2 = 3$$

Now compare coefficients:

$$\alpha = 12b^2 = 12 \cdot 3 = 36$$ $$\beta = 4a^2 = 4 \cdot 1 = 4$$ $$\gamma = 36b^2 - 4a^2 - a^2b^2 = 36\cdot 3 - 4\cdot 1 - 1\cdot 3 = 108 - 4 - 3 = 101$$

Hence,

$$\alpha + \beta + \gamma = 36 + 4 + 101 = 141$$

Therefore, the required value is $141$.