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The curve C has parametric equations $x = 2 \, cos \, t, \; y = \sqrt{3} \; cos \, 2t, \; 0 \leq t \leq \pi$ - Edexcel - A-Level Maths Pure - Question 14 - 2017 - Paper 1
Question 14
The curve C has parametric equations $x = 2 \, cos \, t, \; y = \sqrt{3} \; cos \, 2t, \; 0 \leq t \leq \pi$. (a) Find an expression for \( \frac{dy}{dx} \) in ter... show full transcript
Worked Solution & Example Answer:The curve C has parametric equations $x = 2 \, cos \, t, \; y = \sqrt{3} \; cos \, 2t, \; 0 \leq t \leq \pi$ - Edexcel - A-Level Maths Pure - Question 14 - 2017 - Paper 1
Step 1
Find an expression for \( \frac{dy}{dx} \) in terms of \( t \)
Answer
To find ( \frac{dy}{dx} ), we can use the chain rule:
First, compute ( \frac{dx}{dt} ) and ( \frac{dy}{dt} ):
[ x = 2 \cos(t) \quad \Rightarrow \quad \frac{dx}{dt} = -2 \sin(t) ]
[ y = \sqrt{3} \cos(2t) \quad \Rightarrow \quad \frac{dy}{dt} = -2\sqrt{3} \sin(2t) ]
Now substitute these into the formula:
We can simplify further using the double angle identity, ( \sin(2t) = 2 \sin(t) \cos(t) ):
Step 2
Show that an equation for \( l \) is $2x - 2\sqrt{3}y - 1 = 0$
Answer
To find ( l ), we first find the coordinates of point P when ( t = \frac{2\pi}{3} ):
[ x = 2 \cos \left(\frac{2\pi}{3}\right) = -1, \quad y = \sqrt{3} \cos \left(\frac{4\pi}{3}\right) = -\sqrt{3} ]
Thus, the coordinates of P are ( (-1, -\sqrt{3}) ).
The gradient of the normal line ( l ) is the negative reciprocal of ( \frac{dy}{dx} ) at P:
At ( t = \frac{2\pi}{3} ):
[ \frac{dy}{dx} = 2\sqrt{3} \cdot \cos \left(\frac{2\pi}{3}\right) = 2\sqrt{3} \cdot \left(-\frac{1}{2}\right) = -\sqrt{3} ]
Therefore, the gradient of line ( l ) is:
[ m = \frac{1}{\sqrt{3}} ]
Using point-slope form to find the equation of line ( l ):
[ y - (-\sqrt{3}) = \frac{1}{\sqrt{3}}(x - (-1)) ]
Simplifying:
[ y + \sqrt{3} = \frac{1}{\sqrt{3}}(x + 1) ]
Multiplying through by ( \sqrt{3} ) gives:
[ \sqrt{3}y + 3 = x + 1 \rightarrow 2x - 2\sqrt{3}y - 1 = 0\text{, which is the required form.} ]
Step 3
Find the exact coordinates of Q
Answer
To find the points of intersection, substitute the equation of line ( l ) into the parametric equations:
- Substitute ( y = \frac{1}{\sqrt{3}}(x + 1) - \sqrt{3} ) into the parametric equation of y:
[ y = \sqrt{3} \cos(2t) \rightarrow \frac{1}{\sqrt{3}}(x + 1) - \sqrt{3} = \sqrt{3} \cos(2t) ]
- Using ( x = 2\cos(t) ):
Substituting the expression for ( y ):
[ -2\sqrt{3} \cos(2t) = 2\cos(t) \rightarrow 12\cos^2(t) - 4\cos(t) - 5 = 0 ]
- Solving this quadratic using the quadratic formula:
[ \cos(t) = \frac{4 \pm \sqrt{16 + 240}}{24} = \frac{4 \pm 16}{24} \rightarrow \cos(t) = \frac{5}{6} \text{ or } -\frac{7}{6} ] (discarding the second solution)
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Finding ( t ) from ( \cos(t) = \frac{5}{6} ).
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Substitute back to find ( x ) and ( y ) values:
[ x = 2\left(\frac{5}{6}\right) = \frac{5}{3}, \quad y = \sqrt{3} \cdot \sqrt{1 - \left(\frac{5}{6}\right)^2} = \frac{7}{18\sqrt{3}} ]
Thus, the coordinates of Q are:
[ Q = \left(\frac{5}{3}, \frac{7}{18\sqrt{3}}\right) ]