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 6 - 2013 - Paper 2

To find the points A and B where the line intersects the curve, we substitute the equation of the line into the equation of the ellipse.

1. Substitute $y = x + 2$ into the curve equation: $x^2 + 4(x + 2)^2 - 2x = 35$

2. Expand the equation: $x^2 + 4(x^2 + 4x + 4) - 2x = 35$ $x^2 + 4x^2 + 16x + 16 - 2x = 35$ $5x^2 + 14x + 16 - 35 = 0$ $5x^2 + 14x - 19 = 0$

3. Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 5$, $b = 14$, and $c = -19$:

   $x = \frac{-14 \pm \sqrt{14^2 - 4(5)(-19)}}{2(5)}$ $x = \frac{-14 \pm \sqrt{196 + 380}}{10}$ $x = \frac{-14 \pm \sqrt{576}}{10}$ $x = \frac{-14 \pm 24}{10}$

4. Solve for $x$:

   • First root: $x_1 = \frac{10}{10} = 1$
   • Second root: $x_2 = \frac{-38}{10} = -3.8$

5. Now substitute these $x$ values back into the line equation to find $y$:

   • For $x = 1$: $y_1 = 1 + 2 = 3$
   • For $x = -3.8$: $y_2 = -3.8 + 2 = -1.8$

6. Thus, the coordinates of point A are $(1, 3)$ and the coordinates of point B are $(-3.8, -1.8)$.