A sketch of part of the curve C with equation y = 20 - 4x - \frac{18}{x}, \quad x > 0 is shown in Figure 3. Point A lies on C and has an x coordinate equal to 2. (a) Show that the equation of the normal to C at A is y = -2x + 7. (6) The normal to C at A meets C again at the point B, as shown in Figure 3. (b) Use algebra to find the coordinates of B. (5)

A-Level Edexcel Maths Pure Arithmetic Sequences & Series: A sketch of part of the curve C

A sketch of part of the curve C with equation y = 20 - 4x - \frac{18}{x}, \quad x > 0 is shown in Figure 3 - Edexcel - A-Level Maths Pure - Question 3 - 2014 - Paper 2

Question 3

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A sketch of part of the curve C with equation y = 20 - 4x - \frac{18}{x}, \quad x > 0 is shown in Figure 3. Point A lies on C and has an x coordinate equal to 2. ... show full transcript

Worked Solution & Example Answer:A sketch of part of the curve C with equation y = 20 - 4x - \frac{18}{x}, \quad x > 0 is shown in Figure 3 - Edexcel - A-Level Maths Pure - Question 3 - 2014 - Paper 2

Step 1

Show that the equation of the normal to C at A is y = -2x + 7.

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Answer

To find the equation of the normal to the curve at point A, we need to calculate the derivative of the curve.

Calculate the Derivative: We start by differentiating the equation:

$y = 20 - 4x - \frac{18}{x}$

The derivative is calculated as:

$\frac{dy}{dx} = -4 + \frac{18}{x^2}$
Substitute x = 2: Now substitute x = 2 to find the slope at point A:

$\frac{dy}{dx}\bigg|_{x=2} = -4 + \frac{18}{2^2} = -4 + 4.5 = 0.5$

The slope of the tangent at A is 0.5.
Find the Slope of the Normal: The slope of the normal is the negative reciprocal of the tangent slope:

$m_{normal} = -\frac{1}{0.5} = -2$
Equation of Normal Line: We now use the point-slope form of the line equation. The coordinates of point A when x = 2 are:

$y = 20 - 4(2) - \frac{18}{2} = 20 - 8 - 9 = 3$

Therefore, point A is (2, 3). Using the point-slope form:

$y - y_1 = m(x - x_1)$

Substitute the point (2, 3) and slope -2:

$y - 3 = -2(x - 2)$

Simplifying gives:

$y = -2x + 4 + 3 = -2x + 7$

Thus, we have shown that the normal's equation is indeed:

$y = -2x + 7$

Step 2

Use algebra to find the coordinates of B.

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Answer

To find the coordinates of point B where the normal intersects the curve again:

Set the Equations Equal: We have the normal equation:

$y = -2x + 7$

and the curve's equation:

$y = 20 - 4x - \frac{18}{x}$

Set these equal to each other:

$-2x + 7 = 20 - 4x - \frac{18}{x}$
Rearrange the Equation: Rearranging gives:

$-2x + 4x = 20 - 7 - \frac{18}{x}$

$2x = 13 - \frac{18}{x}$

Multiply through by x to eliminate the fraction:

$2x^2 = 13x - 18$

Rearranging results in:

$2x^2 - 13x + 18 = 0$
Solve the Quadratic Equation: Use the quadratic formula where a = 2, b = -13, c = 18:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substitute the values:

$x = \frac{13 \pm \sqrt{(-13)^2 - 4 \cdot 2 \cdot 18}}{2 \cdot 2}$

$x = \frac{13 \pm \sqrt{169 - 144}}{4}$

$x = \frac{13 \pm 5}{4}$

This gives:

$x = \frac{18}{4} = 4.5$ and $x = \frac{8}{4} = 2$
Find y Coordinates: We already know one coordinate corresponds to A (2, 3). For x = 4.5:

Substitute x = 4.5 into the normal equation:

$y = -2(4.5) + 7 = -9 + 7 = -2$

Thus, the coordinates of point B are:

(4.5, -2)

A sketch of part of the curve C with equation y = 20 - 4x - \frac{18}{x}, \quad x > 0 is shown in Figure 3 - Edexcel - A-Level Maths Pure - Question 3 - 2014 - Paper 2

Worked Solution & Example Answer:A sketch of part of the curve C with equation y = 20 - 4x - \frac{18}{x}, \quad x > 0 is shown in Figure 3 - Edexcel - A-Level Maths Pure - Question 3 - 2014 - Paper 2

Show that the equation of the normal to C at A is y = -2x + 7.

Use algebra to find the coordinates of B.

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