Let's solve the problem step by step. ## 1. **Given Equation** The equation of the two-sheeted hyperboloid is: \[ \left( \frac{x}{a} \right)^2 - \left( \frac{y}{b} \right)^2 - \left( \frac{z}{c} \right)^2 = 1 \] We need to find **a, b, c**, given three points on the surface: - (6, 8, 2) - (3, 4, .5) - (4, √48, ) ## 2. **Plug in each point into the equation** ### **Point 1: (6, 8, 2)** \[ \left( \frac{6}{a} \right)^2 - \left( \frac{8}{b} \right)^2 - \left( \frac{2}{c} \right)^2 = 1 \] \[ \frac{36}{a^2} - \frac{64}{b^2} - \frac{4}{c^2} = 1 \tag{1} \] ### **Point 2: (3, 4, .5)** \[ \left( \frac{3}{a} \right)^2 - \left( \frac{4}{b} \right)^2 - \left( \frac{.5}{c} \right)^2 = 1 \] \[ \frac{9}{a^2} - \frac{16}{b^2} - \frac{.25}{c^2} = 1 \tag{2} \] ### **Point 3: (4, √48, )** \[ \left( \frac{4}{a} \right)^2 - \left( \frac{\sqrt{48}}{b} \right)^2 - \left( \frac{}{c} \right)^2 = 1 \] \[ \frac{16}{a^2} - \frac{48}{b^2} = 1 \tag{3} \] --- ## 3. **Solve for a and b using Equation (3)** \[ \frac{16}{a^2} - \frac{48}{b^2} = 1 \] Try \( a = 4 \) (since \( a \) is a positive integer between 1 and 6): \[ \frac{16}{16} - \frac{48}{b^2} = 1 \implies 1 - \frac{48}{b^2} = 1 \implies \frac{48}{b^2} = \] That doesn't work. Try \( a = 2 \): \[ \frac{16}{4} - \frac{48}{b^2} = 1 \implies 4 - \frac{48}{b^2} = 1 \implies 3 = \frac{48}{b^2} \implies b^2 = 16 \implies b = 4 \] So, **\( a = 2, b = 4 \)**. --- ## 4. **Solve for c using Equation (2) with a = 2, b = 4** \[ \frac{9}{2^2} - \frac{16}{4^2} - \frac{.25}{c^2} = 1 \implies \frac{9}{4} - \frac{16}{16} - \frac{.25}{c^2} = 1 \implies \frac{9}{4} - 1 - \frac{.25}{c^2} = 1 \implies \frac{5}{4} - \frac{.25}{c^2} = 1 \implies \frac{5}{4} - 1 = \frac{.25}{c^2} \implies \frac{1}{4} = \frac{.25}{c^2} \implies c^2 = 1 \implies c = 1 \] --- ## 5. **Final Answer** \[ \boxed{ a = 2 \qquad b = 4 \qquad c = 1 } \] --- ## 6. **Verification with Point 1** Plug \( a = 2, b = 4, c = 1 \) into Equation (1): \[ \frac{36}{4} - \frac{64}{16} - \frac{4}{1} = 9 - 4 - 4 = 1 \] It checks out! --- ## **Final Answers** \[ \boxed{ a = 2,\quad b = 4,\quad c = 1 } \]

# Solution to the Two-Sheeted Hyperboloid Problem ## 1. **Given Equation** The equation of the two-sheeted hyperboloid is: \[ \left( \frac{x}{a} \right)^2 - \left( \frac{y}{b} \right)^2 - \left( \frac{z}{c} \right)^2 = 1 \] We need to find the parameters \( a, b, c \), given the following points on the surface: - \( (6, 8, 2) \) - \( (3, 4, 0.5) \) - \( (4, \sqrt{48}, 0) \) ## 2. **Substituting Points into the Equation** ### **Point 1: \( (6, 8, 2) \)** \[ \left( \frac{6}{a} \right)^2 - \left( \frac{8}{b} \right)^2 - \left( \frac{2}{c} \right)^2 = 1 \] This gives us: \[ \frac{36}{a^2} - \frac{64}{b^2} - \frac{4}{c^2} = 1 \tag{1} \] ### **Point 2: \( (3, 4, 0.5) \)** \[ \left( \frac{3}{a} \right)^2 - \left( \frac{4}{b} \right)^2 - \left( \frac{0.5}{c} \right)^2 = 1 \] This gives us: \[ \frac{9}{a^2} - \frac{16}{b^2} - \frac{0.25}{c^2} = 1 \tag{2} \] ### **Point 3: \( (4, \sqrt{48}, 0) \)** \[ \left( \frac{4}{a} \right)^2 - \left( \frac{\sqrt{48}}{b} \right)^2 = 1 \] This gives us: \[ \frac{16}{a^2} - \frac{48}{b^2} = 1 \tag{3} \] ## 3. **Solving for \( a \) and \( b \) Using Equation (3)** Starting with: \[ \frac{16}{a^2} - \frac{48}{b^2} = 1 \] ### Testing Possible Values for \( a \) - **Try \( a = 4 \)**: \[ \frac{16}{16} - \frac{48}{b^2} = 1 \implies 1 - \frac{48}{b^2} = 1 \implies \text{(no solution)} \] - **Try \( a = 2 \)**: \[ \frac{16}{4} - \frac{48}{b^2} = 1 \implies 4 - \frac{48}{b^2} = 1 \implies 3 = \frac{48}{b^2} \implies b^2 = 16 \implies b = 4 \] Thus, we have: - \( a = 2 \) - \( b = 4 \) ## 4. **Finding \( c \) Using Equation (2)** Now substitute \( a = 2 \) and \( b = 4 \) into Equation (2): \[ \frac{9}{2^2} - \frac{16}{4^2} - \frac{0.25}{c^2} = 1 \] This simplifies to: \[ \frac{9}{4} - 1 - \frac{0.25}{c^2} = 1 \] \[ \frac{5}{4} - \frac{0.25}{c^2} = 1 \] \[ \frac{5}{4} - 1 = \frac{0.25}{c^2} \implies \frac{1}{4} = \frac{0.25}{c^2} \] Solving gives: \[ c^2 = 1 \implies c = 1 \] ## 5. **Final Answer** The parameters are: \[ \boxed{ a = 2, \quad b = 4, \quad c = 1 } \] ## 6. **Verification with Point 1** Substituting \( a = 2, b = 4, c = 1 \) into Equation (1): \[ \frac{36}{4} - \frac{64}{16} - \frac{4}{1} = 9 - 4 - 4 = 1 \] The equation holds true, confirming our values are correct. --- ## **Final Summary** The values of the parameters \( a, b, c \) are: \[ \boxed{ a = 2,\quad b = 4,\quad c = 1 } \]