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Figure 2 shows a sketch of the curve C with parametric equations $x = 1 + t - 5 ext{sin} t,$ $y = 2 - 4 ext{cos} t,$ $- ext{π} < t < ext{π}$ The point A lies on the curve C - Edexcel - A-Level Maths Pure - Question 7 - 2018 - Paper 9
Question 7
Figure 2 shows a sketch of the curve C with parametric equations $x = 1 + t - 5 ext{sin} t,$ $y = 2 - 4 ext{cos} t,$ $- ext{π} < t < ext{π}$ The point A lies on... show full transcript
Worked Solution & Example Answer:Figure 2 shows a sketch of the curve C with parametric equations $x = 1 + t - 5 ext{sin} t,$ $y = 2 - 4 ext{cos} t,$ $- ext{π} < t < ext{π}$ The point A lies on the curve C - Edexcel - A-Level Maths Pure - Question 7 - 2018 - Paper 9
Step 1
find the exact value of k, giving your answer in a fully simplified form.
Answer
To find the value of , we first need to substitute into the equation for :
Setting this equal to 2 gives us:
From this, we can simplify:
Thus, we can find that:
t = rac{ ext{π}}{2}
Next, we substitute this value of back into the equation for :
Setting t = rac{ ext{π}}{2}, we substitute:
x = 1 + rac{ ext{π}}{2} - 5 ext{sin} rac{ ext{π}}{2}
This yields:
x = 1 + rac{ ext{π}}{2} - 5 imes 1 = 1 + rac{ ext{π}}{2} - 5
Thus, reorganizing gives:
k = rac{ ext{π}}{2} - 4
Step 2
Find the equation of the tangent to C at the point A.
Answer
To find the equation of the tangent at point A, we first need to calculate the derivatives of and with respect to :
rac{dx}{dt} = 1 - 5 ext{cos} t rac{dy}{dt} = 4 ext{sin} t
Now we calculate the slope of the tangent line, which is given by:
m = rac{dy/dt}{dx/dt}
Substituting t = rac{ ext{π}}{2} gives us:
m = rac{4 ext{sin}( rac{ ext{π}}{2})}{1 - 5 ext{cos}( rac{ ext{π}}{2})} = rac{4}{1 - 0} = 4
Using the point-slope form of the line, where the point A is and the slope is 4, we derive:
Rearranging leads to:
The equation can thus be expressed in the desired form , where: