The curve $C_1$ with parametric equations $x = 10 ext{cos} \, t, $ $y = 4 ext{√}2 ext{sin} \, t, \, 0 \leq t < 2\pi$ meets the circle $C_2$ with equation $x^2 + y^2 = 66$ at four distinct points as shown in Figure 2 - Edexcel - A-Level Maths Pure - Question 6 - 2019 - Paper 2

The circle $C_2$ has the equation: $x^2 + y^2 = 66$

Now, substitute $y^2 = 66 - x^2$ into the equation for $C_1$:

$\frac{(66 - x^2)}{32} + \frac{x^2}{100} = 1$

Multiply by $3200$ to eliminate the denominators:

$100(66 - x^2) + 32x^2 = 3200$

This simplifies to:

$6600 - 100x^2 + 32x^2 = 3200$

Further simplifying and rearranging gives:

$-68x^2 + 6600 - 3200 = 0\implies 68x^2 = 3400\implies x^2 = 50$

Therefore: $x = \sqrt{50} = 5\text{√}2 \, ext{or} \, -5\text{√}2$

For $x = 5\text{√}2$ and substituting back into the circle equation:

$y^2 = 66 - (5\text{√}2)^2$

Calculating it gives:

$y^2 = 66 - 50 = 16\implies y = 4 \, ext{or} \, -4$

Since we need the 4th quadrant, we take: $y = -4$

Consequently, the Cartesian coordinates for point $S$ are:

$S (5\text{√}2, -4)$

Alternatively, using the other $x$ value: $S = (-5\text{√}2, 4)$, but that is in the 2nd quadrant, so we discard it.