1. Problem 29: Find the value of $k$ such that the line $y = kx + 1$ is tangent to the hyperbola $3x^2 - 4y^2 = 12$. 2. Substitute $y = kx + 1$ into the hyperbola equation: $$3x^2 - 4(kx + 1)^2 = 12$$ 3. Expand the square: $$3x^2 - 4(k^2x^2 + 2kx + 1) = 12$$ 4. Distribute: $$3x^2 - 4k^2x^2 - 8kx - 4 = 12$$ 5. Rearrange terms: $$(3 - 4k^2)x^2 - 8kx - 16 = 0$$ 6. For the line to be tangent, the quadratic in $x$ must have exactly one solution, so the discriminant $D$ must be zero: $$D = (-8k)^2 - 4(3 - 4k^2)(-16) = 0$$ 7. Calculate the discriminant: $$64k^2 + 64(3 - 4k^2) = 0$$ $$64k^2 + 192 - 256k^2 = 0$$ $$-192k^2 + 192 = 0$$ 8. Solve for $k^2$: $$-192k^2 = -192$$ $$k^2 = 1$$ 9. Therefore, $k = \pm 1$. --- 10. Problem 30: Determine the position of point $(7, -3)$ relative to the hyperbola $9x^2 - y^2 = 3$. 11. Substitute $(7, -3)$ into the hyperbola equation: $$9(7)^2 - (-3)^2 = 9 \times 49 - 9 = 441 - 9 = 432$$ 12. Compare with the right side of the equation (3): Since $432 > 3$, the point lies outside the hyperbola. Final answers: - For problem 29, $k = \pm 1$. - For problem 30, the point is outside the hyperbola.