Question 18
The equation of a curve is such that $\dfrac{d^2y}{dx^2} = -\dfrac{24}{x^3}$. It is given that the curve has a stationary point at $(-2, 19)$.
**(d)** Find the equation of the normal to the curve at the point where $\dfrac{dy}{dx} = -\dfrac{9}{4}$ and $x$ is positive. Express your answer in the form $px + qy + r = 0$, where $p$, $q$ and $r$ are integers.
(In parts (a) and (c) you found $\dfrac{dy}{dx} = \dfrac{12}{x^2} - 3$ and $y = -\dfrac{12}{x} - 3x + 7$.)
**(d)** Find the equation of the normal to the curve at the point where $\dfrac{dy}{dx} = -\dfrac{9}{4}$ and $x$ is positive. Express your answer in the form $px + qy + r = 0$, where $p$, $q$ and $r$ are integers.
(In parts (a) and (c) you found $\dfrac{dy}{dx} = \dfrac{12}{x^2} - 3$ and $y = -\dfrac{12}{x} - 3x + 7$.)
