The line $y = x + 2$ meets the curve $x^2 + 4y^2 - 2x = 35$ at the points A and B as shown in Figure 2 - Edexcel - A-Level Maths Pure - Question 2 - 2013 - Paper 3

To find the intersection points A and B of the line and the curve, we first substitute the equation of the line into the curve's equation:

1. Substitute:

   $x^2 + 4(x + 2)^2 - 2x = 35$

2. Expand:

   $x^2 + 4(x^2 + 4x + 4) - 2x = 35$

   $x^2 + 4x^2 + 16 + 16 - 2x = 35$

   $5x^2 + 14x - 27 = 0$

3. Solve the quadratic equation using the quadratic formula:

   $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 5$, $b = 14$, $c = -27$.

   Calculate the discriminant:

   $D = 14^2 - 4 \cdot 5 \cdot (-27) = 196 + 540 = 736$

   Therefore,

   $x = \frac{-14 \pm \sqrt{736}}{10}$

   Since $\sqrt{736} = 8\sqrt{23}$,

   $x = \frac{-14 \pm 8\sqrt{23}}{10}$

   This gives us the two x-coordinates:

4. The coordinates of A and B are derived from:

   $x_A = \frac{-14 + 8\sqrt{23}}{10}, y_A = x_A + 2$

   $x_B = \frac{-14 - 8\sqrt{23}}{10}, y_B = x_B + 2$

5. Finally, we can state the coordinates of A and B as:

   $\text{Coordinates of A: } \left( \frac{-14 + 8\sqrt{23}}{10}, \frac{-14 + 8\sqrt{23}}{10} + 2 \right)$
   $$\text{Coordinates of B: } \left( \frac{-14 - 8\sqrt{23}}{10}, \frac{-14 - 8\sqrt{23}}{10} + 2 \right)$