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The curve C has parametric equations x = 3t - 4, y = 5 - 6t, t > 0 (a) Find \( \frac{dy}{dx} \) in terms of t (b) The point P lies on C where \( t = \frac{1}{2} \) Find the equation of the tangent to C at the point P - Edexcel - A-Level Maths Pure - Question 3 - 2017 - Paper 5
Question 3
The curve C has parametric equations x = 3t - 4, y = 5 - 6t, t > 0 (a) Find \( \frac{dy}{dx} \) in terms of t (b) The point P lies on C where \( t = \frac{1}{2} ... show full transcript
Worked Solution & Example Answer:The curve C has parametric equations x = 3t - 4, y = 5 - 6t, t > 0 (a) Find \( \frac{dy}{dx} \) in terms of t (b) The point P lies on C where \( t = \frac{1}{2} \) Find the equation of the tangent to C at the point P - Edexcel - A-Level Maths Pure - Question 3 - 2017 - Paper 5
Step 1
Find \( \frac{dy}{dx} \) in terms of t
Answer
To find ( \frac{dy}{dx} ), we can use the chain rule:
[
\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}
]
First, we compute ( \frac{dy}{dt} ) and ( \frac{dx}{dt} ):
- For ( y ):
[ \frac{dy}{dt} = \frac{d}{dt}(5 - 6t) = -6 ] - For ( x ):
[ \frac{dx}{dt} = \frac{d}{dt}(3t - 4) = 3 ]
Next, we need ( \frac{dt}{dx} ):
[ \frac{dt}{dx} = \frac{1}{\frac{dx}{dt}} = \frac{1}{3} ] Now substituting into the first equation:
[ \frac{dy}{dx} = -6 \cdot \frac{1}{3} = -2 ]
Step 2
Find the equation of the tangent to C at the point P. Give your answer in the form \( y = px + q \)
Answer
To find the coordinates of point P when ( t = \frac{1}{2} ):
- For ( x ):
[ x = 3 \times \frac{1}{2} - 4 = -2.5 ] - For ( y ):
[ y = 5 - 6 \times \frac{1}{2} = 2 ]
Thus, point P is ( (-2.5, 2) ).
Using the slope found earlier, which is ( -2 ), the equation of the tangent line can be represented as:
[ y - y_1 = m(x - x_1) ]
Substituting the values into point-slope form:
[ y - 2 = -2(x + 2.5) ]
Simplifying gives:
[ y = -2x - 5 + 2 ]
[ y = -2x - 3 ]
Thus, in the form ( y = px + q ), we have ( p = -2 ) and ( q = -3 ).
Step 3
Show that the cartesian equation for C can be written in the form \( y = \frac{ax + b}{x + 4} \)
Answer
Starting from the parametric equations:
[ x = 3t - 4 \implies t = \frac{x + 4}{3} ]
Substituting for ( t ) in ( y = 5 - 6t ):
[ y = 5 - 6\left(\frac{x + 4}{3}\right) ]
The equation simplifies to:
[ y = 5 - 2(x + 4) = 5 - 2x - 8 = -2x - 3 ]
Rearranging in the form provides the equation for C:
[ y = \frac{-2x - 3}{1} ]
Hence, we identify ( a = -2 ) and ( b = -3 ).