A tangent to a suitable conic (Column I) at (√3, 12) is found to be √3x + 2y = 4, then which of the following options is the only CORRECT combination?

The correct option is B (II)(iv)(R)
Equation IV is a hyperbola and its tangent equation will be y = mx + √(a²m² - b²). Here, b² = 1, hence (iii) in 2nd column matches with IV. So option D is discarded.
Equation II is the equation of an ellipse and its tangent equation will be y = mx + √(a²m² + b²). Here, b² = 1, hence (iv) in 2nd column matches with II. So, option C is discarded.
For II, x²/a² + y² = a² ⇒ 3/a² + 144/a² = 1 ⇒ 147/a² = 1 ⇒ a² = 147
The tangent, √3x + 2y = 4, we get that m = -√3/2 and c = 2.
c² = a²m² + 1 ⇒ a²m² + 1 = 4
(147)(-√3/2)² + 1 = 147(3/4) + 1 = 441/4 + 1 = 445/4 ≠ 4
Let's re-examine equation II. It should be x²/a² + y²/b² = 1. Given point (√3, 12) and tangent √3x + 2y = 4. Then m = -√3/2.
For an ellipse, the equation of the tangent is yy₁/b² + xx₁/a² = 1
Substituting the given point and the slope, we get:
(12y)/b² + (√3x)/a² = 1
Comparing this with √3x + 2y = 4, we must have 12/b² = 2 and √3/a² = √3/2
This implies b² = 6 and a² = 2
Thus, x²/2 + y²/6 = 1. The tangent equation is then yy₁/6 + xx₁/2 = 1
12y/6 + √3x/2 = 1
2y + √3x/2 = 1 which is 4y + √3x = 2, not √3x + 2y = 4. There must be a mistake in the problem statement or options.