A sketch of part of the curve C with equation y = 20 - 4x - \frac{18}{x}, \; 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) (b) Use algebra to find the coordinates of B. (5)

A-Level Edexcel Maths Pure Integration: A sketch of part of the curve C

A sketch of part of the curve C with equation y = 20 - 4x - \frac{18}{x}, \; x > 0 is shown in Figure 3 - Edexcel - A-Level Maths Pure - Question 2 - 2013 - Paper 1

Question 2

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

Worked Solution & Example Answer:A sketch of part of the curve C with equation y = 20 - 4x - \frac{18}{x}, \; x > 0 is shown in Figure 3 - Edexcel - A-Level Maths Pure - Question 2 - 2013 - Paper 1

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 at point A, we will first evaluate the function at x = 2:

Substituting for y: $y = 20 - 4(2) - \frac{18}{2} = 20 - 8 - 9 = 3$ Thus, point A is (2, 3).
Finding the derivative: The curve is given by ( y = 20 - 4x - \frac{18}{x} ). The derivative is computed as follows: $\frac{dy}{dx} = -4 + \frac{18}{x^2}$ Substitute ( x = 2 ) into the derivative: $\frac{dy}{dx}\bigg|_{x=2} = -4 + \frac{18}{2^2} = -4 + 4.5 = 0.5$
Calculating the negative reciprocal for the normal line's slope: The slope of the normal line is the negative reciprocal of the derivative: $\text{slope of normal} = -\frac{1}{\frac{dy}{dx}} = -\frac{1}{0.5} = -2$
Using point-slope form to find the equation of the normal: The equation of the normal at point A, using the slope and point (2, 3), is obtained from: $y - y_1 = m(x - x_1)$ $y - 3 = -2(x - 2)$ Thus, $y = -2x + 4 + 3 = -2x + 7$ This confirms the equation of the normal 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 C again:

Setting the equations equal: We know that the normal line's equation is: $y = -2x + 7$ We set this equal to the curve's equation: $-2x + 7 = 20 - 4x - \frac{18}{x}$
Rearranging and simplifying the equation: Bring all terms to one side: $-2x + 7 + 4x + \frac{18}{x} - 20 = 0$ This simplifies to: $2x - 13 + \frac{18}{x} = 0$
Multiplying through by x: To eliminate the fraction, multiply every term by x: $2x^2 - 13x + 18 = 0$
Factoring the quadratic: Factoring gives: $(x - 2)(2x - 9) = 0$ Thus, we have:( x = 2 ) or ( x = \frac{9}{2} ).
Finding the corresponding y-value for the second point B: Substitute ( x = \frac{9}{2} ) back into the equation of the curve: $y = 20 - 4(\frac{9}{2}) - \frac{18}{\frac{9}{2}}$ This simplifies to: $y = 20 - 18 + -4 = -2$
Final coordinates for point B: Thus, the coordinates of point B are ( \left(\frac{9}{2}, -2\right) ).

A sketch of part of the curve C with equation y = 20 - 4x - \frac{18}{x}, \; x > 0 is shown in Figure 3 - Edexcel - A-Level Maths Pure - Question 2 - 2013 - Paper 1

Worked Solution & Example Answer:A sketch of part of the curve C with equation y = 20 - 4x - \frac{18}{x}, \; x > 0 is shown in Figure 3 - Edexcel - A-Level Maths Pure - Question 2 - 2013 - Paper 1

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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