User:Xenocidius/Specialist/Q6

i.
$$ \begin{align} \frac{dt}{dx} & = -\frac{400}{y - 2} \\ \frac{dy}{dx} & = \frac{dy}{dt} \frac{dt}{dx} \\ \therefore \frac{dy}{dx} & = \frac{1}{100} (x - 1) \times -\frac{400}{y - 2} \\ \therefore \frac{dy}{dx} & = -\frac{4(x - 1)}{y - 2} \end{align} $$

ii.
$$ \begin{align} & \therefore (y - 2) \frac{dy}{dx} = -4(x - 1) \\ & \therefore \int \! (y - 2) \frac{dy}{dx} \,dx = \int \! -4(x - 1) \,dx \\ & \therefore \int \! (y - 2) \,dy = \int \! (-4x + 4) \,dx \\ & \therefore \frac{1}{2} y^2 - 2y = -2x^2 + 4x + c \\ \end{align} $$

$$ \begin{align} & \therefore \frac{1}{2} (3)^2 - 2(3) = -2(1)^2 + 4(1) + c \\ & \therefore \frac{9}{2} - 6 = -2 + 4 + c \\ & \therefore \frac{9}{2} - 8 = c \\ & \therefore c = - \frac{7}{2} \\ \end{align} $$

$$ \begin{align} & \therefore \frac{1}{2} y^2 - 2y = -2x^2 + 4x - \frac{7}{2} \\ & \therefore y^2 - 4y = -4x^2 + 8x - 7 \\ & \therefore 4(x^2 - 2x) + y^2 - 4y = -7 \\ & \therefore 4(x^2 - 2x + 1) - 4 + y^2 - 4y + 4 - 4 = -7 \\ & \therefore 4(x - 1)^2 + (y - 2)^2 = 1 \\ & \therefore 2^2(x - 1)^2 + 1^2(y - 2)^2 = 1 \\ & \therefore a = 2, b = 1, h = 1, k = 2 \\ \end{align} $$