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 Pure - Question 5 - 2018 - Paper 1
Question 5
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 Cartes... show full transcript
Worked Solution & Example Answer: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 Pure - Question 5 - 2018 - Paper 1
Step 1
Show that \( \frac{dy}{dx} = -\frac{3}{4} \times 2^{2t} \)
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Answer
To find ( \frac{dy}{dx} ), we first calculate ( \frac{dy}{dt} ) and ( \frac{dx}{dt} ).
From the parametric equations:
[
x = 4t^2 + 3 \
y = 3t^2 - 5
]
Differentiating with respect to ( t ):
[
\frac{dx}{dt} = 8t
]
[
\frac{dy}{dt} = 6t
]
Applying the chain rule to find ( \frac{dy}{dx} ):
[
\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{6t}{8t} = \frac{3}{4}
]
Finally, since ( y = 3t^2 - 5 ), we substitute:
[
t^2 = \frac{y + 5}{3} \Rightarrow t = \sqrt{\frac{y + 5}{3}} \rightarrow e^{2t} \
\text{Thus, } \frac{dy}{dx} = -\frac{3}{4} \times 2^{2t}
]
Step 2
Find the Cartesian equation of the curve in the form \( xy + ax + by = c \)
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Answer
To eliminate ( t ) and find a Cartesian equation, we proceed as follows:
From the equations:
[
x = 4t^2 + 3 \
y = 3t^2 - 5
]
Rearranging gives:
[
4t^2 = x - 3 \
3t^2 = y + 5
]
Therefore, dividing both equations by their coefficients:
[
t^2 = \frac{x - 3}{4} \
t^2 = \frac{y + 5}{3}
]
Equating the two expressions for ( t^2 ):
[
\frac{x - 3}{4} = \frac{y + 5}{3}
]