For a=√2, if a tangent is drawn to a suitable conic (Column I) at the point of contact (√2,1), then which of the following options is the only CORRECT combination for obtaining its equation?

a=√2 and Point of contact : (√2,1)
Comparing with 1st column, x² + y² - a² = 0
(√2,1) = 1 + 1 - (√2)² = 0
∴Equation of curve is x² + y² = a²
∴Equation of tangent = 2x + 2yy' = 0
y' = -x/y
(√2,1) = 1 ⇒ y = -x + √2
Hence, y = -x + √2
Comparing with column 2:

1. my = m²x + a ⇒ Not possible
2. y = mx + a√m²+1 and y = -x + √2
   m = -1 and a√m²+1 = √2
   ∴LHS = RHS
   Hence, option A is correct.