If $x^3 + y^3 = 6$ and $\dfrac{d^2y}{dx^2} \cdot \dfrac{d^2x}{dy^2} = \dfrac{m}{(xy)^n}$, where $m, n \in R$ then $\dfrac{m}{n} = $

If $x = \sin(t), y = \cos(pt)$, then $(1 - x^2)\dfrac{d^2 y}{dx^2} - x\dfrac{dy}{dx} = $

If the function $f(x) = ax^3 + bx^2 + 11x - 6$, defined on $[1, 3]$, satisfies all the conditions of Rolle's theorem for $c = 2 + \dfrac{1}{\sqrt{3}}$, then

The tangent to the curve $y^2 - xy + 9 = 0$ is vertical when