The curve C₁ with parametric equations x = 10cos(t), y = 4√2sin(t), 0 ≤ t < 2π meets the circle C₂ with equation x² + y² = 66 at four distinct points as shown in Figure 2 - Edexcel - A-Level Maths Pure - Question 6 - 2019 - Paper 2

Substituting the expressions for x and y into the equation of the circle: $x² + y² = 66$

Now substituting the y value derived: $x² + \left(4√2√{1 - \left(\frac{x}{10}\right)²}\right)² = 66$

Expanding gives: $x² + 32\left(1 - \left(\frac{x}{10}\right)²\right) = 66$

Rearranging the equation leads to: $x² + 32 - \frac{32x²}{100} = 66$

Combine like terms and multiply through by 100 to clear the fraction: $100x² - 32x² + 3200 - 6600 = 0 \Rightarrow 68x² - 3400 = 0$

Solving for x gives: $x² = 50 \Rightarrow x = ±√50$

Substituting back to find y:

For the fourth quadrant, we take: $x = √{50} = 5√2$

Using: $y = 4√2√{1 - \left(\frac{5√2}{10}\right)²}$

Calculating this step yields: $y = 4√2√{1 - \frac{1}{2}} = 4√2√{\frac{1}{2}} = 4\cdot1\cdot√2 = 2√2$

In the 4th quadrant y must be negative: So the coordinates for S are: S(5√2, -2√2).