A-Level Edexcel Maths Pure Polynomials: The line $y = x + 2$ meets the c

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 1 - 2013 - Paper 2

Question 1

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. (a) Find the coordinates of A and the coordinates of B. (b)... show full transcript

Worked Solution & Example Answer: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 1 - 2013 - Paper 2

Step 1

(a) Find the coordinates of A and the coordinates of B.

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Answer

To find the coordinates of points A and B, we first substitute the equation of the line into the equation of the curve.

Substitution: Substitute $y = x + 2$ into the equation $x^2 + 4y^2 - 2x = 35$ :

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

Expanding this gives:

$x^2 + 4(x^2 + 4x + 4) - 2x = 35$ $x^2 + 4x^2 + 16 + 16x - 2x = 35$ $5x^2 + 14x + 16 - 35 = 0$ $5x^2 + 14x - 19 = 0$
Find x: We now solve this quadratic using the quadratic formula:

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

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

Calculating the two values:
- For $x = \frac{10}{10} = 1$
- For $x = \frac{-38}{10} = -3.8$
Find y coordinates: For $x = 1$ : $y = 1 + 2 = 3$ Thus, point A is $(1, 3)$ .

For $x = -3.8$ : $y = -3.8 + 2 = -1.8$ Thus, point B is $(-3.8, -1.8)$ .

Final coordinates are:

A: $(1, 3)$
B: $(-3.8, -1.8)$

Step 2

(b) Find the distance AB in the form r √2 where r is a rational number.

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Answer

To find the distance between points A and B, we will use the distance formula:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Given points A $(1, 3)$ and B $(-3.8, -1.8)$ :

Calculate x and y differences:

$x_2 - x_1 = -3.8 - 1 = -4.8$ $y_2 - y_1 = -1.8 - 3 = -4.8$
Substitute into the formula:

$d = \sqrt{(-4.8)^2 + (-4.8)^2} = \sqrt{2 \cdot (4.8)^2}$
Simplify:

$= \sqrt{2 \cdot 23.04} = \sqrt{46.08}$ = \sqrt{2 \cdot 23.04} = 4.8 \sqrt{2}, where r = 4.8$$

Thus, the distance AB is $4.8 \sqrt{2}$ .

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 1 - 2013 - Paper 2

Worked Solution & Example Answer: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 1 - 2013 - Paper 2

(a) Find the coordinates of A and the coordinates of B.

(b) Find the distance AB in the form r √2 where r is a rational number.

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