A curve is given by $\frac{dy}{dx} = \frac{2}{a}x^{-\frac{1}{2}} + ax^{-3/2}$, with $a$ a positive constant. The point $A\,(a^2, 3)$ is on this curve.
(i)[3]
Find, in terms of $a$, the equation of the tangent to the curve at $A$, and simplify your answer.
(ii)[4]
Find the equation of the curve.
(iii)[5]
It is also given that $B\,(16, 8)$ lies on the curve. Find the value of $a$ and, using this value, find the distance $AB$.
Worked solution & mark scheme
This 12-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Gradient at $x=a^2$ found as $\frac{dy}{dx}=\frac{3}{a^2}$” …