# Maths A Level Exam Question Help

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Thread starter 1 month ago
#1
Hi, heres the question:

Figure 4 shows a sketch of part of the curve C with equation:
y=32/x^2+3x-8 where x>0
The point P(4,6) lies on C.
The line l is the normal to C at the point P.

The region R, shown shaded in figure 4, is bounded by the line l, the curve C, and the lines with equation x=2 and the x-axis.

Show that the area of R is 46.

How would I approach this question? Im not sure how to attach the sketch of the curve tho 0
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1 month ago
#2
(Original post by hannanj0)
Hi, heres the question:

Figure 4 shows a sketch of part of the curve C with equation:
y=32/x^2+3x-8 where x>0
The point P(4,6) lies on C.
The line l is the normal to C at the point P.

The region R, shown shaded in figure 4, is bounded by the line l, the curve C, and the lines with equation x=2 and the x-axis.

Show that the area of R is 46.

How would I approach this question? Im not sure how to attach the sketch of the curve tho If you quote the post then the Quick Reply box contains a row of icons which include buttons for 'upload image' or 'insert an image via url' or 'insert attachment' depending on how you have taken / stored your image of the sketch 0
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1 month ago
#3
(Original post by hannanj0)
Hi, heres the question:

Figure 4 shows a sketch of part of the curve C with equation:
y=32/x^2+3x-8 where x>0
The point P(4,6) lies on C.
The line l is the normal to C at the point P.

The region R, shown shaded in figure 4, is bounded by the line l, the curve C, and the lines with equation x=2 and the x-axis.

Show that the area of R is 46.

How would I approach this question? Im not sure how to attach the sketch of the curve tho Assuming you’ve been taught differentiation, find the gradient of C at P (and then the equation of the line L) to start with 0
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Thread starter 1 month ago
#4
(Original post by hannanj0)
Hi, heres the question:

Figure 4 shows a sketch of part of the curve C with equation:
y=32/x^2+3x-8 where x>0
The point P(4,6) lies on C.
The line l is the normal to C at the point P.

The region R, shown shaded in figure 4, is bounded by the line l, the curve C, and the lines with equation x=2 and the x-axis.

Show that the area of R is 46.

How would I approach this question? Im not sure how to attach the sketch of the curve tho Here’s the image.
0
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Thread starter 1 month ago
#5
(Original post by laurawatt)
Assuming you’ve been taught differentiation, find the gradient of C at P (and then the equation of the line L) to start with When I did this I got dy/dx= -64x^-3+3, and the equation of the line L as y=2x-2, is this right? Now I'm guessing I will then have to find x when y=0, but not sure after..
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Thread starter 1 month ago
#6
(Original post by davros)
If you quote the post then the Quick Reply box contains a row of icons which include buttons for 'upload image' or 'insert an image via url' or 'insert attachment' depending on how you have taken / stored your image of the sketch I have attached it now 0
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1 month ago
#7
(Original post by hannanj0)
When I did this I got dy/dx= -64x^-3+3, and the equation of the line L as y=2x-2, is this right? Now I'm guessing I will then have to find x when y=0, but not sure after..
Line L is the normal to the point P, you’ve found the equation of the tangent (and you can see on the diagram that line L has a negative gradient!)
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Thread starter 1 month ago
#8
(Original post by laurawatt)
Line L is the normal to the point P, you’ve found the equation of the tangent (and you can see on the diagram that line L has a negative gradient!)
Oops, so now I have y=-0.5x+8, getting an x value of 16. So that would mean the base of the "Triangle" would be 14, now do I need to work out the height, or is that y=6, if not, what do I do then?
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1 month ago
#9
(Original post by hannanj0)
Oops, so now I have y=-0.5x+8, getting an x value of 16. So that would mean the base of the "Triangle" would be 14, now do I need to work out the height, or is that y=6, if not, what do I do then?
My approach would be to find the area of the triangle (from x=4 to x=16) and the area under the curve from x=2 to x=4

As a side note: the height of triangle x=2 to x=16 wouldn’t be 6, that’s the y value of point P 0
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Thread starter 1 month ago
#10
(Original post by laurawatt)
My approach would be to find the area of the triangle (from x=4 to x=16) and the area under the curve from x=2 to x=4

As a side note: the height of triangle x=2 to x=16 wouldn’t be 6, that’s the y value of point P Oh right ok, so how would I obtain the y value?
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1 month ago
#11
(Original post by hannanj0)
Oh right ok, so how would I obtain the y value?
You’ve already been given the y value, (in point P)
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