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

Question

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

UpStudy AI · Accepted Answer

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:
$$
\frac{x^2}{9} - \frac{y^2}{3} = 1
$$

The equation of the hyperbola is: $$ \frac{x^2}{9} - \frac{y^2}{3} = 1 $$

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