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The curve C has parametric equations x = 3t - 4, y = 5 - frac{6}{t}, t > 0 (a) Find \( \frac{dy}{dx} \) in terms of t (2) The point P lies on C where \( t = \tfrac{1}{2} \) (b) Find the equation of the tangent to C at the point P - Edexcel - A-Level Maths Pure - Question 1 - 2017 - Paper 5
Question 1
The curve C has parametric equations x = 3t - 4, y = 5 - frac{6}{t}, t > 0 (a) Find \( \frac{dy}{dx} \) in terms of t (2) The point P lies on C where \( t... show full transcript
Worked Solution & Example Answer:The curve C has parametric equations x = 3t - 4, y = 5 - frac{6}{t}, t > 0 (a) Find \( \frac{dy}{dx} \) in terms of t (2) The point P lies on C where \( t = \tfrac{1}{2} \) (b) Find the equation of the tangent to C at the point P - Edexcel - A-Level Maths Pure - Question 1 - 2017 - Paper 5
Step 1
Find \( \frac{dy}{dx} \) in terms of t
Answer
To find ( \frac{dy}{dx} ), we will use the chain rule. First, we need to find ( \frac{dy}{dt} ) and ( \frac{dx}{dt} ).
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Differentiate ( y = 5 - \frac{6}{t} ): [ \frac{dy}{dt} = 0 + 6t^{-2} = \frac{6}{t^2} ]
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Differentiate ( x = 3t - 4 ): [ \frac{dx}{dt} = 3 ]
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Now apply the chain rule: [ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\frac{6}{t^2}}{3} = \frac{2}{t^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 equation of the tangent line, we first need to determine the coordinates of point P when ( t = \frac{1}{2} ).
- Calculate the coordinates:
- For ( x ): [ x = 3(\frac{1}{2}) - 4 = \frac{3}{2} - 4 = -\frac{5}{2} ]
- For ( y ): [ y = 5 - \frac{6}{\frac{1}{2}} = 5 - 12 = -7 ]
Thus, the coordinates of point P are ( \left(-\frac{5}{2}, -7\right) ).
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Now substitute ( t = \frac{1}{2} ) into ( \frac{dy}{dx} ): [ \frac{dy}{dx} = \frac{2}{(\frac{1}{2})^2} = \frac{2}{\frac{1}{4}} = 8 ]
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The slope of the tangent line is 8. Using point-slope form: [ y - y_1 = m(x - x_1) ] [ y + 7 = 8\left(x + \frac{5}{2}\right) ] [ y = 8x + 20 - 7 = 8x + 13 ]
Thus, in the form ( y = px + q ), we have ( p = 8 ) and ( q = 13 ).
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 this into the equation for y:
- [ y = 5 - \frac{6}{t} = 5 - \frac{6}{\frac{x + 4}{3}} = 5 - \frac{18}{x + 4} ]
Now simplifying further:
- [ y = \frac{5(x + 4) - 18}{x + 4} = \frac{5x + 20 - 18}{x + 4} = \frac{5x + 2}{x + 4} ]
Thus, in the form ( y = \frac{ax + b}{x + 4} ), we can see that ( a = 5 ) and ( b = 2 ).