A coordinate sketch of the curve x = 16t², y = t(t² − 9) for t > 0, with horizontal and vertical axes and origin O.

The diagram shows the curve given parametrically by the equations $x = 16t^2$ , $y = t(t^2 - 9)$ , for $t > 0$

Show that $dydx=3t2−932t\displaystyle \frac{dy}{dx} = \frac{3t^2 - 9}{32t}$

Find the coordinates of the point on the curve at which the tangent to the curve is parallel to the line $3 x - 16 y + 8 = 0$

Find the cartesian equation of the curve.

A coordinate sketch of the curve x = 16t², y = t(t² − 9) for t > 0, with horizontal and vertical axes and origin O.

The diagram shows the curve given parametrically by the equations $x = 16t^2$ , $y = t(t^2 - 9)$ , for $t > 0$

Show that $dydx=3t2−932t\displaystyle \frac{dy}{dx} = \frac{3t^2 - 9}{32t}$

Find the coordinates of the point on the curve at which the tangent to the curve is parallel to the line $3 x - 16 y + 8 = 0$

Find the cartesian equation of the curve.

Differentiation

Differentiation Questions