Question

If the equation of the hyperbola with foci \( (4, 2) \) and \( (8, 2) \) is \( 3x^2 - y^2 - \alpha x + \beta y + \gamma = 0 \), then \( \alpha + \beta + \gamma \) is equal to _______

Hint:

Find the center and the value of \( c \) from the foci. Use the standard equation of the hyperbola with the major axis parallel to the x-axis. Use the relation \( c^2 = a^2 + b^2 \). Compare the coefficients of the given equation with the expanded form of the standard equation to find \( a^2, b^2, \alpha, \beta, \gamma \).

Answer 1

Correct Answer: 141

The foci of the hyperbola are \( S'(4, 2) \) and \( S(8, 2) \). 
The center of the hyperbola is the midpoint of the foci: \[ C = \left( \frac{4 + 8}{2}, \frac{2 + 2}{2} \right) = (6, 2) \] The distance between the foci is \( 2c = \sqrt{(8 - 4)^2 + (2 - 2)^2} = \sqrt{4^2 + 0^2} = 4 \), so \( c = 2 \). 
The major axis is parallel to the x-axis since the y-coordinates of the foci are the same. 
The equation of the hyperbola is of the form \( \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \), where \( (h, k) \) is the center. 
Here, \( (h, k) = (6, 2) \), so the equation is \( \frac{(x - 6)^2}{a^2} - \frac{(y - 2)^2}{b^2} = 1 \). 
We have the relation \( c^2 = a^2 + b^2 \), so \( 2^2 = a^2 + b^2 \Rightarrow 4 = a^2 + b^2 \Rightarrow b^2 = 4 - a^2 \). 
Since the equation involves \( y^2 \) term with a negative coefficient, we assume the given form is obtained after some manipulation of the standard equation. 
Let's consider the possibility that the order of terms was swapped in the standard equation, leading to \( \frac{(y - 2)^2}{a^2} - \frac{(x - 6)^2}{b^2} = 1 \). 
In this case, the major axis would be parallel to the y-axis, which contradicts the foci coordinates. So the first form is correct. 
From the given equation \( 3x^2 - y^2 - \alpha x + \beta y + \gamma = 0 \), we can rewrite the standard equation: 

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

Comparing with \( 3x^2 - y^2 - \alpha x + \beta y + \gamma = 0 \), we can assume a scaling factor \( k \): \( kb^2 = 3 \) 

\( -ka^2 = -1 \Rightarrow ka^2 = 1 \) \( -12kb^2 = -\alpha \Rightarrow \alpha = 12kb^2 = 12(3) = 36 \) 

\( 4ka^2 = \beta \Rightarrow \beta = 4(1) = 4 \) \( k(36b^2 - 4a^2 - a^2 b^2) = \gamma \Rightarrow 36(3) - 4(1) - (1)(3) = \gamma \Rightarrow 108 - 4 - 3 = \gamma \Rightarrow \gamma = 101 \) We have \( b^2 = 4 - a^2 \). 

From \( ka^2 = 1 \) and \( kb^2 = 3 \), we get \( \frac{b^2}{a^2} = 3 \Rightarrow b^2 = 3a^2 \). So, \( 3a^2 = 4 - a^2 \Rightarrow 4a^2 = 4 \Rightarrow a^2 = 1 \). 

Then \( b^2 = 3a^2 = 3(1) = 3 \). Now, \( \alpha = 36 \), \( \beta = 4 \), \( \gamma = 101 \). \( \alpha + \beta + \gamma = 36 + 4 + 101 = 141 \).