Figure 3 shows a sketch of part of the curve C with equation y = \frac{1}{2}x^2 + 27 - 12, \text{where } x > 0. The point A lies on C and has coordinates \left( 3, -\frac{3}{2} \right). (a) Show that the equation of the normal to C at A can be written as 10y = 4x - 27. (5) (b) The normal to C at A meets C again at the point B, as shown in Figure 3. Use algebra to find the coordinates of B.

A-Level Edexcel Maths Pure Vectors in 3 Dimensions: Figure 3 shows a sketch of part

Figure 3 shows a sketch of part of the curve C with equation y = \frac{1}{2}x^2 + 27 - 12, \text{where } x > 0 - Edexcel - A-Level Maths Pure - Question 2 - 2018 - Paper 2

Question 2

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Figure 3 shows a sketch of part of the curve C with equation y = \frac{1}{2}x^2 + 27 - 12, \text{where } x > 0. The point A lies on C and has coordinates \left( 3... 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^2 + 27 - 12, \text{where } x > 0 - Edexcel - A-Level Maths Pure - Question 2 - 2018 - Paper 2

Step 1

Show that the equation of the normal to C at A can be written as 10y = 4x - 27

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Answer

To find the equation of the normal at point A, we first need to determine the derivative of the curve at A to get the slope of the tangent.

Find the derivative of the curve: Given the curve equation: $y = \frac{1}{2}x^2 + 27 - 12$ We differentiate: $\frac{dy}{dx} = x$ Evaluating at ( x = 3 ): $\frac{dy}{dx} \bigg|_{x=3} = 3$
Calculate the slope of the normal: The slope of the tangent at point A is 3, therefore: $m_{normal} = -\frac{1}{m_{tangent}} = -\frac{1}{3}$
Using point-slope form: The point-slope form of a line is: $y - y_1 = m(x - x_1)$ Plugging in the coordinates of A ( (3, -\frac{3}{2}) ) and the normal slope: $y + \frac{3}{2} = -\frac{1}{3}(x - 3)$
Simplifying the equation: Multiply through by 3 to eliminate the fraction: $3y + 3(\frac{3}{2}) = -(x - 3)$ $3y + \frac{9}{2} = -x + 3$ Rearranging gives: $3y = -x + 3 - \frac{9}{2}$ $3y = -x - \frac{3}{2}$ Multiplying by -1: $-3y = x + \frac{3}{2}$ This can be re-arranged to: $10y = 4x - 27$

Step 2

Use algebra to find the coordinates of B

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Answer

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

Set the two equations equal: Substitute ( y = \frac{1}{2}x^2 + 27 - 12 ) into the equation of the normal:

Start with the normal equation: $10y = 4x - 27$
Substituting for y: $10(\frac{1}{2}x^2 + 15) = 4x - 27$ Simplifying gives: $5x^2 + 150 = 4x - 27$ Thus: $5x^2 - 4x + 177 = 0$
Applying the quadratic formula: Using the quadratic formula ( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ), we identify:
- ( a = 5 )
- ( b = -4 )
- ( c = 177 ) The value of the discriminant ( b^2 - 4ac ) can be calculated: $\Delta = (-4)^2 - 4(5)(177) = 16 - 3540 = -3524$ Since the discriminant is negative, there are no real solutions, indicating the normal does not intersect the curve again for x > 0.

The curve does not have a second intersection point with the normal; therefore, we need a different approach or to check calculations as errors might have occurred.

Figure 3 shows a sketch of part of the curve C with equation y = \frac{1}{2}x^2 + 27 - 12, \text{where } x > 0 - Edexcel - A-Level Maths Pure - Question 2 - 2018 - Paper 2

Worked Solution & Example Answer:Figure 3 shows a sketch of part of the curve C with equation y = \frac{1}{2}x^2 + 27 - 12, \text{where } x > 0 - Edexcel - A-Level Maths Pure - Question 2 - 2018 - Paper 2

Show that the equation of the normal to C at A can be written as 10y = 4x - 27

Use algebra to find the coordinates of B

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