A-Level Edexcel Maths Pure Trigonometric Functions: Figure 2 shows a sketch of the c

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 6 - 2018 - Paper 9

Question 6

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 6-2018-Paper 9.png

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 li... 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 6 - 2018 - Paper 9

Step 1

find the exact value of k, giving your answer in a fully simplified form.

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Answer

To find the value of k, we start from the equation for y:

2 - 4 ext{ cos } t = 2

Solving this gives:

4 ext{ cos } t = 0 ightarrow ext{cos } t = 0

The solutions for cos t = 0 within the range $-\pi < t < \pi$ are:

t = \frac{\pi}{2}, \quad -\frac{\pi}{2}

Next, we substitute these t values back into the equation for x to find k:

For $t = \frac{\pi}{2}$ :

k = 1 + \frac{\pi}{2} - 5 \text{ sin } \frac{\pi}{2} = 1 + \frac{\pi}{2} - 5$$

k = -4 + \frac{\pi}{2}

For $t = -\frac{\pi}{2}$:

k = 1 - \frac{\pi}{2} - 5 \text{ sin } -\frac{\pi}{2} = 1 - \frac{\pi}{2} + 5$$

k = 6 - \frac{\pi}{2}

Since k > 0, we take:

k = 6 - \frac{\pi}{2}

Step 2

Find the equation of the tangent to C at the point A.

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Answer

To find the equation of the tangent line, we need the derivative of y with respect to t:

\frac{dy}{dt} = 4 \text{ sin } t \\ \frac{dx}{dt} = 1 - 5 \text{ cos } t

The slope of the tangent line (m) can be found using:

m = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4 \text{ sin } t}{1 - 5 \text{ cos } t}

At the point A, when $t = \frac{\pi}{2}$ :

\frac{dy}{dt} = 4 \\ \frac{dx}{dt} = 1 - 5(0) = 1

So the slope m is:

m = \frac{4}{1} = 4

The equation of the tangent line in point-slope form is:

y - 2 = 4(x - (6 - \frac{\pi}{2}))

Rearranging gives:

y = 4x - 24 + 2\frac{\pi}{2} + 2$$ Now we simplify:

y = 4x - 22 + \pi

Thus, in the form $y = px + q$, we have:

p = 4, \ q = -22 + \pi$$

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 6 - 2018 - Paper 9

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 6 - 2018 - Paper 9

find the exact value of k, giving your answer in a fully simplified form.

Find the equation of the tangent to C at the point A.

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