The equation of the common tangent with positive slope to the parabola y²= 8√3x and the hyperbola 4x²–y² = 4 is(A) y = √6 x + √2 (B) y = √6 x – √2 (C) y = √3 x + √2 (D) y = √3 x – √2

Let the equation of tangent be y=mx+cEquation of parabola : y² = 8√3xCondition of tangency for parabola is c=a/mwhere a is coefficient of x divide by 4Here a=8√3÷4Which is equal to 2√3 Therefore , c=2√3÷m..................(1)Now for Hyperbola 4x² - y²=4Or x²/1-y²/4=1Condition of tangency for hyperbola is c=(+,-)√a²m²-b²For this hyperbola a² = 1 and b²=4Therefore c=(+,-)√1²m²-4...........................(2)LHS of equation (1) and (2) are equal so there RHS must be equal Therefore, 2√3÷m=(+,-)√m²-4On solving this we get only one real solution that is m=√6Putting m=√6 in equation (1)We get c=√2 Therefore the required equation of tangent is y=√6x+√2