A-Level Edexcel Maths Pure Vectors in 3 Dimensions: Figure 3 shows a sketch of the c

Figure 3 shows a sketch of the curve C with parametric equations $x = 4 \, ext{cos} \left( t + \frac{\pi}{6} \right)$, $y = 2 \, \text{sin} \, t$, $0 < t < 2\pi$ (a) Show that $x + y = 2 \sqrt{3} \, \text{cos} \, t$ (b) Show that a cartesian equation of C is $(x + y)^2 + a y^2 = b$ where a and b are integers to be determined. - Edexcel - A-Level Maths Pure - Question 8 - 2014 - Paper 7

Question 8

$Figure-3-shows-a-sketch-of-the-curve-C-with-parametric-equations--$x-=-4-\,--ext{cos}-\left(-t-+-\frac{\pi}{6}-\right)$,--$y-=-2-\,-\text{sin}-\,-t$,--$0-<-t-<-2\pi$--(a)-Show-that--$x-+-y-=-2-\sqrt{3}-\,-\text{cos}-\,-t$--(b)-Show-that-a-cartesian-equation-of-C-is--$(x-+-y)^2-+-a-y^2-=-b$--where-a-and-b-are-integers-to-be-determined.-Edexcel-A-Level Maths Pure-Question 8-2014-Paper 7.png$

Figure 3 shows a sketch of the curve C with parametric equations $x = 4 \, ext{cos} \left( t + \frac{\pi}{6} \right)$, $y = 2 \, \text{sin} \, t$, $0 < t < 2\pi$... show full transcript

Worked Solution & Example Answer:Figure 3 shows a sketch of the curve C with parametric equations $x = 4 \, ext{cos} \left( t + \frac{\pi}{6} \right)$, $y = 2 \, \text{sin} \, t$, $0 < t < 2\pi$ (a) Show that $x + y = 2 \sqrt{3} \, \text{cos} \, t$ (b) Show that a cartesian equation of C is $(x + y)^2 + a y^2 = b$ where a and b are integers to be determined. - Edexcel - A-Level Maths Pure - Question 8 - 2014 - Paper 7

Step 1

Show that $x + y = 2 \sqrt{3} \, \text{cos} \, t$

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Answer

To begin, we will express both $x$ and $y$ in terms of $t$ :

From the equation for $x$ :
$x = 4 \text{cos} \left( t + \frac{\pi}{6} \right)$
From the equation for $y$ :
$y = 2 \text{sin} \, t$

Using the angle addition formula: $\text{cos} \left( t + \frac{\pi}{6} \right) = \text{cos} \, t \cdot \text{cos} \left( \frac{\pi}{6} \right) - \text{sin} \, t \cdot \text{sin} \left( \frac{\pi}{6} \right)$ Substituting the values,\ \text{cos} \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2} ext{ and } \text{sin} \left( \frac{\pi}{6} \right) = \frac{1}{2}$:

$x = 4 \left( \frac{\sqrt{3}}{2} \text{cos} \, t - \frac{1}{2} \text{sin} \, t \right)$

This simplifies to:

$x = 2\sqrt{3} \text{cos} \, t - 2 \text{sin} \, t$

Combining with the equation for $y$ : $x + y = 2\sqrt{3} \text{cos} \, t - 2 \text{sin} \, t + 2\text{sin} \, t$

So, $x + y = 2\sqrt{3} \text{cos} \, t$

Thus, we have shown that $x + y = 2 \sqrt{3} \, \text{cos} \, t$ .

Step 2

Show that a cartesian equation of C is $(x + y)^2 + a y^2 = b$

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Answer

First, we start from the derived equation:

$x + y = 2 \sqrt{3} \text{cos} \, t$

Square both sides: $(x + y)^2 = (2 \sqrt{3} \text{cos} \, t)^2$ $= 12 \text{cos}^2 \, t$

Now, using the identity $\text{sin}^2 \, t + \text{cos}^2 \, t = 1$ : From the equation $y = 2\text{sin} \, t$ , we express $\text{sin} \, t$ as: $\text{sin} \, t = \frac{y}{2}$ Thus, we have:
$\text{cos}^2 \, t = 1 - \left(\frac{y}{2}\right)^2 = \frac{4 - y^2}{4}$

Now substitute $\text{cos}^2 \, t$ into our earlier equation:

$(x + y)^2 = 12 \cdot \frac{4 - y^2}{4} = 3(4 - y^2) = 12 - 3y^2$

Rearranging gives: $(x + y)^2 + 3y^2 = 12$

Thus, we conclude that a Cartesian equation of C is: $(x + y)^2 + 3y^2 = 12$

Identifying $a = 3$ and $b = 12$ , where $a$ and $b$ are integers.

Show that $x + y = 2 \sqrt{3} \, \text{cos} \, t$

Show that a cartesian equation of C is $(x + y)^2 + a y^2 = b$

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