# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2017 | Oct-Nov | (P1-9709/12) | Q#10

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Question

The diagram shows part of the curve  and the normal to the curve at the point P(2, 3).  This normal meets the x-axis at Q.

i.       Find the equation of the normal at P.

ii.       Find, showing all necessary working, the area of the shaded region.

Solution

i.

We are required to find the equation of the normal to the curve at point P(2,3).

To find the equation of the line either we need coordinates of the two points on the line (Two-Point  form of Equation of Line) or coordinates of one point on the line and slope of the line (Point-Slope  form of Equation of Line).

We already have coordinates of a point on the normal P(2,3).

Therefore, we need slope of the normal to the curve.

If a line  is normal to the curve , then product of their slopes  and  at that point (where line  is normal to the curve) is;

Therefore, we can find slope of the normal by finding slope (gradient) of the curve at point P(2,3)  where normal meets the curve.

Gradient (slope) of the curve at the particular point is the derivative of equation of the curve at that  particular point.

Gradient (slope)  of the curve  at a particular point  can be found by substituting x- coordinates of that point in the expression for gradient of the curve;

We are given equation of the curve as;

Therefore;

Rule for differentiation of  is:

Now we find gradient of the curve at point P(2,3).

Hence;

Now we can write equation of the normal.

Point-Slope form of the equation of the line is;

ii.

Consider the diagram below.

It is evident from the diagram that;

We can see that;

Let’s first find area A.

To find the area of region under the curve , we need to integrate the curve from point  to  along x-axis.

We are given equation of the curve as;

Therefore;

It is evident from the diagram that area under curve extends along x-axis from x-intercept of curve  to intersection point of curve and the normal.

We need to find x-intercept of the curve first.

The point  at which curve (or line) intercepts x-axis, the value of . So we can find the  value of  coordinate by substituting  in the equation of the curve (or line).

We are given equation of the curve as;

Substitute ;

Next we need x-coordinates of the point of intersection of curve and the normal. We are given that point P(2,3) is intersection point of the curve and the normal.

Hence;

Rule for integration of  is:

Next we need to find area B.

Expression for the area of the triangle is;

For the given case;

It is evident from the diagram that height of the triangle PQR is PR which is equal to y-coordinates  of point P(2,3). Therefore;

It can also be seen from the diagram that base of triangle PQR is QR which extends from x-coordinate of point P(2,3) to x-intercept of normal to the curve.

Therefore, we need to find x-intercept of the normal.

The point  at which curve (or line) intercepts x-axis, the value of . So we can find the  value of  coordinate by substituting  in the equation of the curve (or line).

We have found equation of normal to the curve in (i) as;

Substitute ;

Therefore, QR, base of triangle PQR;

Hence;

Finally, we can find the area of shaded region;