A curve is defined by the parametric equations $x = 4t^2 + 3$ y = 3t^2 - 5 5 (a) Show that $$\frac{dy}{dx} = -\frac{3}{4} \times 2^{2t}$$ 5 (b) Find the Cartesian equation of the curve in the form $xy + ax + by = c$, where $a$, $b$, and $c$ are integers. - AQA - A-Level Maths Mechanics - Question 5 - 2018 - Paper 1

To show that ( \frac{dy}{dx} = -\frac{3}{4} \times 2^{2t} ), we start by differentiating the parametric equations.

1. Differentiate y and x with respect to t:

   $\frac{dy}{dt} = \frac{d}{dt}(3t^2 - 5) = 6t$

   $\frac{dx}{dt} = \frac{d}{dt}(4t^2 + 3) = 8t$

2. Use the chain rule to find $\frac{dy}{dx}$:

   $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{6t}{8t} = \frac{6}{8} = \frac{3}{4}$

3. Substituting for t:

   • We know that from the equation of x, we can rearrange to find (t):

   $t = \sqrt{\frac{x - 3}{4}}$

   • Now substituting into the expression for $\frac{dy}{dx}$, we get:

   $\frac{dy}{dx} = -\frac{3}{4} \times 2^{2t}$