The diagram shows the curve given parametrically by the equations $x = kt^2$ , $y = t(t^2 - 8)$ , for $t > 0$ , where $k\,$ is a positive constant

Show that $dydx=3t2−82kt\displaystyle \frac{dy}{dx} = \frac{3t^2 - 8}{2kt}$

Find the coordinates, in terms of $k$ , of the point on the curve at which the tangent to the curve is parallel to the line $3 x - 12 y + 8 = 0$

Find the cartesian equation of the curve, in terms of $k$ .

The diagram shows the curve given parametrically by the equations $x = kt^2$ , $y = t(t^2 - 8)$ , for $t > 0$ , where $k\,$ is a positive constant

Show that $dydx=3t2−82kt\displaystyle \frac{dy}{dx} = \frac{3t^2 - 8}{2kt}$

Find the coordinates, in terms of $k$ , of the point on the curve at which the tangent to the curve is parallel to the line $3 x - 12 y + 8 = 0$

Find the cartesian equation of the curve, in terms of $k$ .

Differentiation

Differentiation Questions