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 problem asks for the value of the sum \( \alpha + \beta + \gamma \) given the foci and the general equation of a hyperbola.

Concept Used:

The solution involves comparing the properties of a hyperbola derived from its standard equation with the properties derived from its given general equation.

1. Properties from Foci: For a hyperbola with foci \(S_1(x_1, y_1)\) and \(S_2(x_2, y_2)\):

• The center of the hyperbola \((h, k)\) is the midpoint of the segment connecting the foci: \( h = \frac{x_1+x_2}{2}, k = \frac{y_1+y_2}{2} \).
• The distance between the foci is \(2c\).
• The orientation of the transverse axis depends on whether the x or y coordinates of the foci are constant. If the y-coordinates are the same, the transverse axis is horizontal.

2. Standard Equation: The standard equation of a hyperbola with a horizontal transverse axis and center \((h, k)\) is:

\[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \]

3. Relationship between parameters: For any hyperbola, \(c^2 = a^2 + b^2\).

4. General Equation to Standard Form: A general second-degree equation can be converted to the standard form by completing the square for the x and y terms.

Step-by-Step Solution:

Step 1: Determine the properties of the hyperbola from its foci.

The given foci are \( S_1(4, 2) \) and \( S_2(8, 2) \).

The center \((h, k)\) is the midpoint of the foci:

\[ h = \frac{4 + 8}{2} = 6 \] \[ k = \frac{2 + 2}{2} = 2 \]

So, the center of the hyperbola is C(6, 2).

The distance between the foci is \(2c\):

\[ 2c = \sqrt{(8-4)^2 + (2-2)^2} = \sqrt{4^2} = 4 \] \[ c = 2 \implies c^2 = 4 \]

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

Step 2: Convert the given general equation to the standard form.

The given equation is \( 3x^2 - y^2 - \alpha x + \beta y + \gamma = 0 \).

We group the x and y terms and complete the square:

\[ (3x^2 - \alpha x) - (y^2 - \beta y) + \gamma = 0 \] \[ 3\left(x^2 - \frac{\alpha}{3}x\right) - \left(y^2 - \beta y\right) + \gamma = 0 \] \[ 3\left(x^2 - \frac{\alpha}{3}x + \left(\frac{\alpha}{6}\right)^2\right) - 3\left(\frac{\alpha}{6}\right)^2 - \left(y^2 - \beta y + \left(\frac{\beta}{2}\right)^2\right) + \left(\frac{\beta}{2}\right)^2 + \gamma = 0 \] \[ 3\left(x - \frac{\alpha}{6}\right)^2 - \left(y - \frac{\beta}{2}\right)^2 = \frac{3\alpha^2}{36} - \frac{\beta^2}{4} - \gamma \] \[ 3\left(x - \frac{\alpha}{6}\right)^2 - \left(y - \frac{\beta}{2}\right)^2 = \frac{\alpha^2}{12} - \frac{\beta^2}{4} - \gamma \]

To get the standard form \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), we divide by the right-hand side. Let \(K = \frac{\alpha^2}{12} - \frac{\beta^2}{4} - \gamma\). Then:

\[ \frac{(x - \alpha/6)^2}{K/3} - \frac{(y - \beta/2)^2}{K} = 1 \]

Step 3: Compare the properties from both forms to find \(\alpha\) and \(\beta\).

By comparing the center \((h, k) = (\alpha/6, \beta/2)\) from the equation with the center \((6, 2)\) from the foci:

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

Step 4: Find the value of \(\gamma\).

From the standard form derived from the general equation, we have:

\[ a^2 = \frac{K}{3} \quad \text{and} \quad b^2 = K \]

Using the relation \(c^2 = a^2 + b^2\) and the value \(c^2=4\):

\[ 4 = \frac{K}{3} + K \implies 4 = \frac{4K}{3} \implies K = 3 \]

Now, substitute the values of \(K\), \(\alpha\), and \(\beta\) into the expression for \(K\):

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

Step 5: Calculate the final sum \( \alpha + \beta + \gamma \).

We have found \(\alpha = 36\), \(\beta = 4\), and \(\gamma = 101\).

\[ \alpha + \beta + \gamma = 36 + 4 + 101 = 141 \]

The value of \( \alpha + \beta + \gamma \) is 141.

Answer 2

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, \( \alpha = 36 \), \( \beta = 4 \), \( \gamma = 101 \). \( \alpha + \beta + \gamma = 36 + 4 + 101 = 141 \).