From point (2, 2) tangents are drawn to the hyperbola $\frac{x^2}{16} - \frac{y^2}{9} = 1$ then point of contact lies in
Step-by-Step Solution
Key Concept: For a point P(2,2) external to hyperbola x²/16 - y²/9 = 1, the equation of chord of contact is (2x)/16 - (2y)/9 = 1, which simplifies to 9x - 8y = 36. The points of contact satisfy both this chord equation and the hyperbola equation simultaneously.
For the hyperbola with asymptotes $4y - 3x = 0$ and $4y + 3x = 0$, tangents from external point $(2, 2)$ are found using the condition that the point lies above both asymptotes. The contact points of tangents drawn from $(2, 2)$ lie in the third and fourth quadrants where the hyperbola exists relative to these asymptotes, as verified by substituting into the asymptotic equations.
Correct Answer: 3,4