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Figure 3 shows a sketch of part of the curve C with equation y = \frac{1}{2}x + 27 - 12, \quad x > 0 The point A lies on C and has coordinates \( \left( 3, -\frac{3}{2} \right) \) - Edexcel - A-Level Maths Pure - Question 1 - 2018 - Paper 1
Question 1
Figure 3 shows a sketch of part of the curve C with equation y = \frac{1}{2}x + 27 - 12, \quad x > 0 The point A lies on C and has coordinates \( \left( 3, -\frac{... show full transcript
Worked Solution & Example Answer:Figure 3 shows a sketch of part of the curve C with equation y = \frac{1}{2}x + 27 - 12, \quad x > 0 The point A lies on C and has coordinates \( \left( 3, -\frac{3}{2} \right) \) - Edexcel - A-Level Maths Pure - Question 1 - 2018 - Paper 1
Step 1
a) Show that the equation of the normal to C at A can be written as 10y = 4x - 27.
Answer
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Find the derivative of the curve C:
Given the equation of the curve, we take the derivative:
[\frac{dy}{dx} = \frac{1}{2} (1) + 0 - 0 = \frac{1}{2}\text{ when }x = 3.] -
Evaluate the gradient at point A:
Substitute (x = 3) into (\frac{dy}{dx}):
[m_t = \frac{1}{2}] -
Find the negative reciprocal (slope of normal):
[m_{normal} = -\frac{1}{m_t} = -2.] -
Use point-slope form to write the equation of the normal:
Using the point (A(3, -\frac{3}{2})):
[y - (-\frac{3}{2}) = -2(x - 3) ]
Simplifying gives us:
[y + \frac{3}{2} = -2x + 6 ]
[y = -2x + 6 - \frac{3}{2} = -2x + \frac{12}{2} - \frac{3}{2} = -2x + \frac{9}{2}.] -
Rearranging gives:
[10y = 4x -27.]
Thus, we have shown that the equation of the normal is correctly expressed as (10y = 4x - 27).
Step 2
b) Use algebra to find the coordinates of B.
Answer
-
Substituting the equation of the normal into that of the curve:
Set the two equations equal to each other to find intersection points. We have:
[y = \frac{1}{2}x + 27 - 12 \quad\text{and}\quad 10y = 4x - 27.] -
Substituting for y in the curve's equation:
Replace y in the curve's equation:
[10\left(\frac{1}{2}x + 15\right) = 4x - 27.]
This simplifies to:
[5x + 150 = 4x - 27.] -
Bringing like terms together:
[5x - 4x = -27 - 150.]
[x = -177.] -
Finding y using x value:
Substitute back into either equation, for example, the curve:
[y = \frac{1}{2}(-177) + 15.]
[y = -88.5 + 15 = -73.5.]
Thus, the coordinates of point B are ((-177, -73.5)).