# Maths A Level Exam Question Help

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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

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

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#2

(Original post by

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

**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

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#3

**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

0

reply

**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.

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(Original post by

Assuming you’ve been taught differentiation, find the gradient of C at P (and then the equation of the line L) to start with

**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

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(Original post by

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

**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

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#7

(Original post by

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..

**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..

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(Original post by

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!)

**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!)

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#9

(Original post by

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?

**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?

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

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(Original post by

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

**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

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#11

(Original post by

Oh right ok, so how would I obtain the y value?

**hannanj0**)Oh right ok, so how would I obtain the y value?

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