\( \begin{array}{ll}\text { 280. Zapisite enačbo hiperbole s sredičcem } S(0,0) \text { in } & \text { a) Nap } \\ \text { gorisci na abscisni osi, ki gre skozi tocki } A \text { in } B, & \text { b) Ira } \\ \begin{array}{ll}\text { a) } A(6,3), B(2 \sqrt{3}, 1) \text { b) } A(-5,4), B(3 \sqrt{2},-3) & \text { e) gor } \\ & \end{array}\end{array} \)

Alright, I need to translate and solve the given problem about hyperbola equations with the center at \( S(0,0) \). The problem is presented in Slovenian, so I'll first translate it into English to understand it better. **Problem Translation:** "280. Write the equation of a hyperbola with center \( S(0,0) \) and: a) Vertices on the x-axis passing through points \( A \) and \( B \), b) Ira (This part seems incomplete or unclear, possibly a typo or missing information)." **Clarifying the Problem:** It seems like the problem is asking to write the equation of a hyperbola centered at the origin with vertices on the x-axis, passing through specific points \( A \) and \( B \). The second part, labeled "b) Ira," appears to be incomplete or possibly a typo. I'll focus on solving part (a) as it is clear. **Understanding the Hyperbola Equation:** A standard hyperbola centered at the origin with vertices on the x-axis has the equation: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \] where \( a \) is the distance from the center to each vertex along the x-axis. **Given Points:** - \( A(6,3) \) - \( B(2\sqrt{3},1) \) **Objective:** Determine the values of \( a \) and \( b \) such that the hyperbola passes through points \( A \) and \( B \). **Approach:** 1. **Substitute Point A into the Hyperbola Equation:** \[ \frac{6^2}{a^2} - \frac{3^2}{b^2} = 1 \implies \frac{36}{a^2} - \frac{9}{b^2} = 1 \quad \text{(Equation 1)} \] 2. **Substitute Point B into the Hyperbola Equation:** \[ \frac{(2\sqrt{3})^2}{a^2} - \frac{1^2}{b^2} = 1 \implies \frac{12}{a^2} - \frac{1}{b^2} = 1 \quad \text{(Equation 2)} \] 3. **Solve the System of Equations:** - From Equation 1: \( \frac{36}{a^2} - \frac{9}{b^2} = 1 \) - From Equation 2: \( \frac{12}{a^2} - \frac{1}{b^2} = 1 \) Let's denote \( \frac{1}{a^2} = A \) and \( \frac{1}{b^2} = B \) to simplify: \[ 36A - 9B = 1 \quad \text{(Equation 1a)} \] \[ 12A - B = 1 \quad \text{(Equation 2a)} \] 4. **Solve for \( A \) and \( B \):** - From Equation 2a: \( B = 12A - 1 \) - Substitute \( B \) into Equation 1a: \[ 36A - 9(12A - 1) = 1 \implies 36A - 108A + 9 = 1 \implies -72A = -8 \implies A = \frac{8}{72} = \frac{1}{9} \] - Now, find \( B \): \[ B = 12 \times \frac{1}{9} - 1 = \frac{12}{9} - 1 = \frac{4}{3} - 1 = \frac{1}{3} \] 5. **Determine \( a \) and \( b \):** - \( A = \frac{1}{a^2} = \frac{1}{9} \implies a^2 = 9 \implies a = 3 \) - \( B = \frac{1}{b^2} = \frac{1}{3} \implies b^2 = 3 \implies b = \sqrt{3} \) 6. **Write the Equation of the Hyperbola:** \[ \frac{x^2}{3^2} - \frac{y^2}{(\sqrt{3})^2} = 1 \implies \frac{x^2}{9} - \frac{y^2}{3} = 1 \] **Conclusion:** The equation of the hyperbola with center at \( S(0,0) \), vertices on the x-axis, and passing through points \( A(6,3) \) and \( B(2\sqrt{3},1) \) is: \[ \frac{x^2}{9} - \frac{y^2}{3} = 1 \]